Maths-

General

Easy

Question

$3 tan to the power of 6 space 10 to the power of ring operator minus 27 tan to the power of 4 space 10 to the power of ring operator plus 33 tan squared space 10 to the power of ring operator space$ =

0
1
2
3

Hint:

In this question, we have too find the value of the given equation. For this we will use the value of tan 30 and later simplify it in the required way to find the value.

The correct answer is: 1

$3 tan to the power of 6 space 10 to the power of ring operator minus 27 tan to the power of 4 space 10 to the power of ring operator plus 33 tan squared space 10 to the power of ring operator A s comma space tan space 30 degree equals fraction numerator 1 over denominator square root of 3 end fraction rightwards double arrow tan 3 left parenthesis 10 degree right parenthesis equals fraction numerator 1 over denominator square root of 3 end fraction space rightwards double arrow fraction numerator left parenthesis 3 tan 10 degree minus tan cubed 10 degree right parenthesis over denominator left parenthesis 1 minus 3 tan squared 10 degree right parenthesis end fraction equals fraction numerator 1 over denominator square root of 3 end fraction S q u a r i n g space b o t h space s i d e s comma space w e space g e t colon rightwards double arrow left parenthesis 3 tan 10 degree minus tan cubed 10 degree right parenthesis squared over left parenthesis 1 minus 3 tan squared 10 degree right parenthesis squared equals 1 third rightwards double arrow fraction numerator 9 tan squared 10 degree minus 6 tan to the power of 4 10 degree plus tan to the power of 6 10 degree over denominator 1 minus 6 tan squared 10 degree plus 9 tan to the power of 4 10 degree end fraction equals 1 third rightwards double arrow 27 tan squared 10 degree minus 18 tan to the power of 4 10 degree plus 3 tan to the power of 6 10 degree equals 1 minus 6 tan squared 10 degree plus 9 tan to the power of 4 10 degree rightwards double arrow 3 tan to the power of 6 space 10 to the power of ring operator minus 27 tan to the power of 4 space 10 to the power of ring operator plus 33 tan squared space 10 to the power of ring operator equals 1$

$sin space 20 to the power of ring operator open parentheses 4 plus sec space 20 to the power of ring operator close parentheses$ =

space space space space space sin space 20 to the power of ring operator open parentheses 4 plus sec space 20 to the power of ring operator close parentheses equals 4 sin space 20 degree plus fraction numerator sin space 20 degree over denominator cos space 20 degree end fraction equals fraction numerator 4 sin 20 degree. cos 20 degree plus sin space 20 degree over denominator cos space 20 degree end fraction equals fraction numerator 2 left parenthesis 2 sin space 20 degree. cos space 20 degree right parenthesis plus sin space 20 degree over denominator cos space 20 degree end fraction equals fraction numerator 2 sin space 40 degree plus sin space 20 degree over denominator cos space 20 degree end fraction equals fraction numerator 2 sin space left parenthesis 60 minus 20 right parenthesis degree plus sin space 20 degree over denominator cos space 20 degree end fraction equals fraction numerator 2 open parentheses sin space 60 degree cos space 20 degree minus cos space 60 degree sin space 20 degree close parentheses plus sin space 20 degree over denominator cos space 20 degree end fraction equals fraction numerator 2 open parentheses begin display style fraction numerator square root of 3 over denominator 2 end fraction end style cos space 20 degree minus begin display style 1 half end style sin space 20 degree close parentheses plus sin space 20 degree over denominator cos space 20 degree end fraction equals fraction numerator square root of 3 cos space 20 degree minus sin space 20 degree plus sin space 20 degree over denominator cos space 20 degree end fraction equals fraction numerator square root of 3 cos space 20 degree over denominator cos space 20 degree end fraction equals square root of 3

$sin space 20 to the power of ring operator open parentheses 4 plus sec space 20 to the power of ring operator close parentheses$ =

Maths-General

space space space space space sin space 20 to the power of ring operator open parentheses 4 plus sec space 20 to the power of ring operator close parentheses equals 4 sin space 20 degree plus fraction numerator sin space 20 degree over denominator cos space 20 degree end fraction equals fraction numerator 4 sin 20 degree. cos 20 degree plus sin space 20 degree over denominator cos space 20 degree end fraction equals fraction numerator 2 left parenthesis 2 sin space 20 degree. cos space 20 degree right parenthesis plus sin space 20 degree over denominator cos space 20 degree end fraction equals fraction numerator 2 sin space 40 degree plus sin space 20 degree over denominator cos space 20 degree end fraction equals fraction numerator 2 sin space left parenthesis 60 minus 20 right parenthesis degree plus sin space 20 degree over denominator cos space 20 degree end fraction equals fraction numerator 2 open parentheses sin space 60 degree cos space 20 degree minus cos space 60 degree sin space 20 degree close parentheses plus sin space 20 degree over denominator cos space 20 degree end fraction equals fraction numerator 2 open parentheses begin display style fraction numerator square root of 3 over denominator 2 end fraction end style cos space 20 degree minus begin display style 1 half end style sin space 20 degree close parentheses plus sin space 20 degree over denominator cos space 20 degree end fraction equals fraction numerator square root of 3 cos space 20 degree minus sin space 20 degree plus sin space 20 degree over denominator cos space 20 degree end fraction equals fraction numerator square root of 3 cos space 20 degree over denominator cos space 20 degree end fraction equals square root of 3

General

Maths-

In the figure $P Q greater than P R comma$ QS and RS are the bisectors of $stack ∣ Q with _ below$ and $stack R with _ below$ respectively then

PQ>PR
=

angle

PRQ>

angle

PQR [angle opposite to the longer side of triangle is greater]
=1/2

angle

PRQ> 1/2

angle

PQR
=

angle

SRQ>

angle

SQR [as QR is the bisector of

angle

PQR and RS is the bisector of

angle

PRQ]
= SQ>SR [as side opposite to greater angle is greatest]
So, SQ>SR

In the figure $P Q greater than P R comma$ QS and RS are the bisectors of $stack ∣ Q with _ below$ and $stack R with _ below$ respectively then

Maths-General

PQ>PR
=

angle

PRQ>

angle

PQR [angle opposite to the longer side of triangle is greater]
=1/2

angle

PRQ> 1/2

angle

PQR
=

angle

SRQ>

angle

SQR [as QR is the bisector of

angle

PQR and RS is the bisector of

angle

PRQ]
= SQ>SR [as side opposite to greater angle is greatest]
So, SQ>SR

General

Maths-

If AB=AC and $A C D equals 115 to the power of ring operator end exponent$ then $stack B with _ below A C equals$

Maths-General

General

Maths-

In the figure $A equals 30 to the power of ring operator end exponent & E B equals 50 to the power of ring operator end exponent$ then ACD=

Maths-General

General

Maths-

$open parentheses sin to the power of 8 space 75 to the power of ring operator minus cos to the power of 8 space 75 to the power of ring operator close parentheses$ =

space space space open parentheses sin to the power of 8 space 75 to the power of ring operator minus cos to the power of 8 space 75 to the power of ring operator close parentheses equals sin to the power of 8 open parentheses 90 minus 15 close parentheses degree minus cos to the power of 8 left parenthesis 90 minus 15 right parenthesis degree equals cos to the power of 8 15 degree minus sin to the power of 8 15 degree equals open parentheses cos to the power of 4 15 degree plus sin to the power of 4 15 degree close parentheses open parentheses cos to the power of 4 15 degree minus sin to the power of 4 15 degree close parentheses equals open parentheses open parentheses cos squared 15 degree plus sin squared 15 degree close parentheses squared minus 2 sin squared 15 degree. cos squared space 15 degree close parentheses open parentheses cos squared 15 degree plus sin squared 15 degree close parentheses open parentheses cos squared 15 degree minus sin squared 15 degree close parentheses equals open parentheses 1 minus open parentheses 2 sin space 15 degree. cos 15 degree close parentheses squared over 2 close parentheses 1 cross times cos open parentheses 2 cross times 15 close parentheses degree equals open parentheses 1 minus fraction numerator s i n squared 30 degree over denominator 2 end fraction close parentheses cos space 30 degree equals open parentheses 1 minus 1 over 8 close parentheses fraction numerator square root of 3 over denominator 2 end fraction equals 7 over 8. fraction numerator square root of 3 over denominator 2 end fraction equals fraction numerator 7 square root of 3 over denominator 16 end fraction

$open parentheses sin to the power of 8 space 75 to the power of ring operator minus cos to the power of 8 space 75 to the power of ring operator close parentheses$ =

Maths-General

space space space open parentheses sin to the power of 8 space 75 to the power of ring operator minus cos to the power of 8 space 75 to the power of ring operator close parentheses equals sin to the power of 8 open parentheses 90 minus 15 close parentheses degree minus cos to the power of 8 left parenthesis 90 minus 15 right parenthesis degree equals cos to the power of 8 15 degree minus sin to the power of 8 15 degree equals open parentheses cos to the power of 4 15 degree plus sin to the power of 4 15 degree close parentheses open parentheses cos to the power of 4 15 degree minus sin to the power of 4 15 degree close parentheses equals open parentheses open parentheses cos squared 15 degree plus sin squared 15 degree close parentheses squared minus 2 sin squared 15 degree. cos squared space 15 degree close parentheses open parentheses cos squared 15 degree plus sin squared 15 degree close parentheses open parentheses cos squared 15 degree minus sin squared 15 degree close parentheses equals open parentheses 1 minus open parentheses 2 sin space 15 degree. cos 15 degree close parentheses squared over 2 close parentheses 1 cross times cos open parentheses 2 cross times 15 close parentheses degree equals open parentheses 1 minus fraction numerator s i n squared 30 degree over denominator 2 end fraction close parentheses cos space 30 degree equals open parentheses 1 minus 1 over 8 close parentheses fraction numerator square root of 3 over denominator 2 end fraction equals 7 over 8. fraction numerator square root of 3 over denominator 2 end fraction equals fraction numerator 7 square root of 3 over denominator 16 end fraction

General

Maths-

If $tan space alpha equals fraction numerator 1 over denominator square root of x open parentheses x squared plus x plus 1 close parentheses end root end fraction comma tan space beta equals fraction numerator square root of x over denominator square root of x squared plus x plus 1 end root end fraction$ and $tan space gamma equals square root of x to the power of negative 3 end exponent plus x to the power of negative 2 end exponent plus x to the power of negative 1 end exponent end root$ then $alpha plus beta$ =

tan space alpha equals fraction numerator 1 over denominator square root of x open parentheses x squared plus x plus 1 close parentheses end root end fraction comma space tan space beta equals fraction numerator square root of x over denominator square root of x squared plus x plus 1 end root end fraction space a n d space tan space gamma equals square root of x to the power of negative 3 end exponent plus x to the power of negative 2 end exponent plus x to the power of negative 1 end exponent end root space space space space tan space left parenthesis alpha plus beta right parenthesis equals fraction numerator tan space alpha plus tan space beta over denominator 1 minus tan space alpha. tan space beta end fraction equals fraction numerator fraction numerator 1 over denominator square root of x open parentheses x squared plus x plus 1 close parentheses end root end fraction plus fraction numerator square root of x over denominator square root of x squared plus x plus 1 end root end fraction over denominator 1 minus fraction numerator 1 over denominator square root of x open parentheses x squared plus x plus 1 close parentheses end root end fraction. fraction numerator square root of x over denominator square root of x squared plus x plus 1 end root end fraction end fraction equals fraction numerator square root of x squared plus x plus 1 end root plus x square root of x squared plus x plus 1 end root over denominator square root of x left parenthesis x squared plus x plus 1 right parenthesis minus square root of x end fraction equals fraction numerator open parentheses x plus 1 close parentheses square root of x squared plus x plus 1 end root over denominator square root of x open parentheses x squared plus x plus 1 minus 1 close parentheses end fraction equals fraction numerator left parenthesis x plus 1 right parenthesis square root of x squared plus x plus 1 end root over denominator x square root of x open parentheses x plus 1 close parentheses end fraction equals fraction numerator square root of x squared plus x plus 1 end root over denominator square root of x cubed end root end fraction equals square root of fraction numerator x squared plus x plus 1 over denominator x cubed end fraction end root equals square root of 1 over x plus 1 over x squared plus 1 over x cubed end root equals square root of x to the power of negative 3 end exponent plus x to the power of negative 2 end exponent plus x to the power of negative 1 end exponent end root equals tan space gamma A s comma space tan space left parenthesis alpha plus beta right parenthesis equals tan space gamma rightwards double arrow alpha plus beta equals gamma

If $tan space alpha equals fraction numerator 1 over denominator square root of x open parentheses x squared plus x plus 1 close parentheses end root end fraction comma tan space beta equals fraction numerator square root of x over denominator square root of x squared plus x plus 1 end root end fraction$ and $tan space gamma equals square root of x to the power of negative 3 end exponent plus x to the power of negative 2 end exponent plus x to the power of negative 1 end exponent end root$ then $alpha plus beta$ =

Maths-General

tan space alpha equals fraction numerator 1 over denominator square root of x open parentheses x squared plus x plus 1 close parentheses end root end fraction comma space tan space beta equals fraction numerator square root of x over denominator square root of x squared plus x plus 1 end root end fraction space a n d space tan space gamma equals square root of x to the power of negative 3 end exponent plus x to the power of negative 2 end exponent plus x to the power of negative 1 end exponent end root space space space space tan space left parenthesis alpha plus beta right parenthesis equals fraction numerator tan space alpha plus tan space beta over denominator 1 minus tan space alpha. tan space beta end fraction equals fraction numerator fraction numerator 1 over denominator square root of x open parentheses x squared plus x plus 1 close parentheses end root end fraction plus fraction numerator square root of x over denominator square root of x squared plus x plus 1 end root end fraction over denominator 1 minus fraction numerator 1 over denominator square root of x open parentheses x squared plus x plus 1 close parentheses end root end fraction. fraction numerator square root of x over denominator square root of x squared plus x plus 1 end root end fraction end fraction equals fraction numerator square root of x squared plus x plus 1 end root plus x square root of x squared plus x plus 1 end root over denominator square root of x left parenthesis x squared plus x plus 1 right parenthesis minus square root of x end fraction equals fraction numerator open parentheses x plus 1 close parentheses square root of x squared plus x plus 1 end root over denominator square root of x open parentheses x squared plus x plus 1 minus 1 close parentheses end fraction equals fraction numerator left parenthesis x plus 1 right parenthesis square root of x squared plus x plus 1 end root over denominator x square root of x open parentheses x plus 1 close parentheses end fraction equals fraction numerator square root of x squared plus x plus 1 end root over denominator square root of x cubed end root end fraction equals square root of fraction numerator x squared plus x plus 1 over denominator x cubed end fraction end root equals square root of 1 over x plus 1 over x squared plus 1 over x cubed end root equals square root of x to the power of negative 3 end exponent plus x to the power of negative 2 end exponent plus x to the power of negative 1 end exponent end root equals tan space gamma A s comma space tan space left parenthesis alpha plus beta right parenthesis equals tan space gamma rightwards double arrow alpha plus beta equals gamma

General

Maths-

If $tan space left parenthesis pi cos space theta right parenthesis equals cot space left parenthesis pi sin space theta right parenthesis comma 0 less than theta less than fraction numerator 3 pi over denominator 4 end fraction$ then $sin space open parentheses theta plus pi over 4 close parentheses equals$

space space space space tan left parenthesis pi space cos space ϑ right parenthesis equals c o t space left parenthesis pi space sin space ϑ right parenthesis rightwards double arrow tan left parenthesis pi space cos space ϑ right parenthesis equals tan space open parentheses pi over 2 minus pi space sin space ϑ close parentheses rightwards double arrow pi space cos space ϑ equals pi over 2 minus pi space sin space ϑ rightwards double arrow cos space ϑ equals 1 half minus sin space ϑ rightwards double arrow cos space ϑ plus sin space ϑ equals 1 half M u l t i p l y i n g space b o t h space t h e space s i d e s space w i t h space fraction numerator 1 over denominator square root of 2 end fraction comma space w e space g e t colon rightwards double arrow fraction numerator 1 over denominator square root of 2 end fraction cos space ϑ plus fraction numerator 1 over denominator square root of 2 end fraction sin space ϑ equals fraction numerator 1 over denominator 2 square root of 2 end fraction rightwards double arrow sin space pi over 4 cos space ϑ plus cos space pi over 4 sin space ϑ equals fraction numerator 1 over denominator 2 square root of 2 end fraction rightwards double arrow sin open parentheses ϑ plus pi over 4 close parentheses equals fraction numerator 1 over denominator 2 square root of 2 end fraction

If $tan space left parenthesis pi cos space theta right parenthesis equals cot space left parenthesis pi sin space theta right parenthesis comma 0 less than theta less than fraction numerator 3 pi over denominator 4 end fraction$ then $sin space open parentheses theta plus pi over 4 close parentheses equals$

Maths-General

space space space space tan left parenthesis pi space cos space ϑ right parenthesis equals c o t space left parenthesis pi space sin space ϑ right parenthesis rightwards double arrow tan left parenthesis pi space cos space ϑ right parenthesis equals tan space open parentheses pi over 2 minus pi space sin space ϑ close parentheses rightwards double arrow pi space cos space ϑ equals pi over 2 minus pi space sin space ϑ rightwards double arrow cos space ϑ equals 1 half minus sin space ϑ rightwards double arrow cos space ϑ plus sin space ϑ equals 1 half M u l t i p l y i n g space b o t h space t h e space s i d e s space w i t h space fraction numerator 1 over denominator square root of 2 end fraction comma space w e space g e t colon rightwards double arrow fraction numerator 1 over denominator square root of 2 end fraction cos space ϑ plus fraction numerator 1 over denominator square root of 2 end fraction sin space ϑ equals fraction numerator 1 over denominator 2 square root of 2 end fraction rightwards double arrow sin space pi over 4 cos space ϑ plus cos space pi over 4 sin space ϑ equals fraction numerator 1 over denominator 2 square root of 2 end fraction rightwards double arrow sin open parentheses ϑ plus pi over 4 close parentheses equals fraction numerator 1 over denominator 2 square root of 2 end fraction

General

Maths-

The minimum value $m space equals bold space 2 cos space theta plus fraction numerator 1 over denominator sin invisible function application theta end fraction plus square root of 2 tan space theta space text in end text open parentheses 0 comma pi over 2 close parentheses$ is

m equals 2 space cos space ϑ plus fraction numerator 1 over denominator sin space ϑ end fraction plus square root of 2 tan space ϑ sin c e comma space ϑ element of open parentheses 0 comma pi over 2 close parentheses S o comma space sin space ϑ greater than 0 comma space cos space ϑ greater than 0 comma tan space ϑ greater than 0 u sin g space A M thin space greater or equal than G M space p r o p e r t y. rightwards double arrow fraction numerator 2 space cos space ϑ plus begin display style fraction numerator 1 over denominator sin space ϑ end fraction end style plus square root of 2 tan space ϑ over denominator 3 end fraction greater or equal than 3 open parentheses 2 square root of 2 close parentheses to the power of bevelled 1 third end exponent rightwards double arrow 2 space cos space ϑ plus fraction numerator 1 over denominator sin space ϑ end fraction plus square root of 2 tan space ϑ greater or equal than 3 open parentheses square root of 2 cubed close parentheses to the power of bevelled 1 third end exponent rightwards double arrow 2 space cos space ϑ plus fraction numerator 1 over denominator sin space ϑ end fraction plus square root of 2 tan space ϑ greater or equal than 3 square root of 2 S o comma space m i n i m u m space v a l u e space i s space 3 square root of 2.

The minimum value $m space equals bold space 2 cos space theta plus fraction numerator 1 over denominator sin invisible function application theta end fraction plus square root of 2 tan space theta space text in end text open parentheses 0 comma pi over 2 close parentheses$ is

Maths-General

m equals 2 space cos space ϑ plus fraction numerator 1 over denominator sin space ϑ end fraction plus square root of 2 tan space ϑ sin c e comma space ϑ element of open parentheses 0 comma pi over 2 close parentheses S o comma space sin space ϑ greater than 0 comma space cos space ϑ greater than 0 comma tan space ϑ greater than 0 u sin g space A M thin space greater or equal than G M space p r o p e r t y. rightwards double arrow fraction numerator 2 space cos space ϑ plus begin display style fraction numerator 1 over denominator sin space ϑ end fraction end style plus square root of 2 tan space ϑ over denominator 3 end fraction greater or equal than 3 open parentheses 2 square root of 2 close parentheses to the power of bevelled 1 third end exponent rightwards double arrow 2 space cos space ϑ plus fraction numerator 1 over denominator sin space ϑ end fraction plus square root of 2 tan space ϑ greater or equal than 3 open parentheses square root of 2 cubed close parentheses to the power of bevelled 1 third end exponent rightwards double arrow 2 space cos space ϑ plus fraction numerator 1 over denominator sin space ϑ end fraction plus square root of 2 tan space ϑ greater or equal than 3 square root of 2 S o comma space m i n i m u m space v a l u e space i s space 3 square root of 2.

General

Maths-

In the adjoining figure, $stack left floor a with _ below equals left floor b$ and $left floor c equals left floor d comma$ then $left floor b plus left floor c equals$

Maths-General

General

Maths-

In the figure, $a equals 4 x to the power of ring operator end exponent$ and $b equals left parenthesis 2.5 x plus 24 right parenthesis to the power of ring operator end exponent comma$ then the value of x =

Maths-General

General

Maths-

In the following figure, the value of x =

Maths-General

General

Maths-

Using information given in the following figure, the value of x and y is

Maths-General

General

Maths-

In figure, the sides QP and RQ of $triangle P Q R$ are produced to points S and T respectively. If $stack S P R with _ below equals 135 to the power of ring operator end exponent$ and $open floor P Q T equals 110 to the power of ring operator end exponent close$ then $stack P R Q with _ below equals$

Maths-General

General

Maths-

In the given figure, if $P Q R equals 70 to the power of ring operator end exponent$ then $∣ A B C equals$

PQ is parallel to AB and QR is the transversal
Also, QR is parallel to BC and AB is the transversal.

angle A B C equals angle Q O B equals angle P Q R

70 degree

As angles are alternate interior angles so the angles are equal.

In the given figure, if $P Q R equals 70 to the power of ring operator end exponent$ then $∣ A B C equals$

Maths-General

PQ is parallel to AB and QR is the transversal
Also, QR is parallel to BC and AB is the transversal.

angle A B C equals angle Q O B equals angle P Q R

70 degree

As angles are alternate interior angles so the angles are equal.

General

Maths-

From the given figure, the value of ‘a’ is

As the AB is parallel to CD and there is transversal
So, the vertical opposite angle of

50 degree

as those are always equal.
Now, we have the interior angles, and we know that their sum is

180 degree

S o comma 50 degree plus a equals 180 degree rightwards double arrow a equals 180 degree minus 50 degree rightwards double arrow a equals 130 degree

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Can’t find what you’re looking for?

$3 tan to the power of 6 space 10 to the power of ring operator minus 27 tan to the power of 4 space 10 to the power of ring operator plus 33 tan squared space 10 to the power of ring operator space$ =

0
1
2
3

In this question, we have too find the value of the given equation. For this we will use the value of tan 30 and later simplify it in the required way to find the value.

The correct answer is: 1

Related Questions to study

$sin space 20 to the power of ring operator open parentheses 4 plus sec space 20 to the power of ring operator close parentheses$ =

$sin space 20 to the power of ring operator open parentheses 4 plus sec space 20 to the power of ring operator close parentheses$ =

In the figure $P Q greater than P R comma$ QS and RS are the bisectors of $stack ∣ Q with _ below$ and $stack R with _ below$ respectively then

In the figure $P Q greater than P R comma$ QS and RS are the bisectors of $stack ∣ Q with _ below$ and $stack R with _ below$ respectively then

If AB=AC and $A C D equals 115 to the power of ring operator end exponent$ then $stack B with _ below A C equals$

If AB=AC and $A C D equals 115 to the power of ring operator end exponent$ then $stack B with _ below A C equals$

In the figure $A equals 30 to the power of ring operator end exponent & E B equals 50 to the power of ring operator end exponent$ then ACD=

In the figure $A equals 30 to the power of ring operator end exponent & E B equals 50 to the power of ring operator end exponent$ then ACD=

$open parentheses sin to the power of 8 space 75 to the power of ring operator minus cos to the power of 8 space 75 to the power of ring operator close parentheses$ =

$open parentheses sin to the power of 8 space 75 to the power of ring operator minus cos to the power of 8 space 75 to the power of ring operator close parentheses$ =

The minimum value $m space equals bold space 2 cos space theta plus fraction numerator 1 over denominator sin invisible function application theta end fraction plus square root of 2 tan space theta space text in end text open parentheses 0 comma pi over 2 close parentheses$ is

The minimum value $m space equals bold space 2 cos space theta plus fraction numerator 1 over denominator sin invisible function application theta end fraction plus square root of 2 tan space theta space text in end text open parentheses 0 comma pi over 2 close parentheses$ is

In the adjoining figure, $stack left floor a with _ below equals left floor b$ and $left floor c equals left floor d comma$ then $left floor b plus left floor c equals$

In the adjoining figure, $stack left floor a with _ below equals left floor b$ and $left floor c equals left floor d comma$ then $left floor b plus left floor c equals$

In the figure, $a equals 4 x to the power of ring operator end exponent$ and $b equals left parenthesis 2.5 x plus 24 right parenthesis to the power of ring operator end exponent comma$ then the value of x =

In the figure, $a equals 4 x to the power of ring operator end exponent$ and $b equals left parenthesis 2.5 x plus 24 right parenthesis to the power of ring operator end exponent comma$ then the value of x =

In the following figure, the value of x =

In the following figure, the value of x =

Using information given in the following figure, the value of x and y is

Using information given in the following figure, the value of x and y is

In the given figure, if $P Q R equals 70 to the power of ring operator end exponent$ then $∣ A B C equals$

In the given figure, if $P Q R equals 70 to the power of ring operator end exponent$ then $∣ A B C equals$

From the given figure, the value of ‘a’ is

From the given figure, the value of ‘a’ is

Have a query? Ask a Tutor!!

Can’t find what you’re looking for?

=

0123

In this question, we have too find the value of the given equation. For this we will use the value of tan 30 and later simplify it in the required way to find the value.

The correct answer is: 1

Related Questions to study

=

=

In the figure QS and RS are the bisectors of and respectively then

In the figure QS and RS are the bisectors of and respectively then

If AB=AC and then

If AB=AC and then

In the figure then ACD=

In the figure then ACD=

=

=

If and then =

If and then =

If then

If then

The minimum value is

The minimum value is

In the adjoining figure, and then

In the adjoining figure, and then

In the figure, and then the value of x =

In the figure, and then the value of x =

In the following figure, the value of x =

In the following figure, the value of x =

Using information given in the following figure, the value of x and y is

Using information given in the following figure, the value of x and y is

In figure, the sides QP and RQ of are produced to points S and T respectively. If and then

In figure, the sides QP and RQ of are produced to points S and T respectively. If and then

In the given figure, if then

In the given figure, if then

From the given figure, the value of ‘a’ is

From the given figure, the value of ‘a’ is

$3 tan to the power of 6 space 10 to the power of ring operator minus 27 tan to the power of 4 space 10 to the power of ring operator plus 33 tan squared space 10 to the power of ring operator space$ =

0
1
2
3

$sin space 20 to the power of ring operator open parentheses 4 plus sec space 20 to the power of ring operator close parentheses$ =

$sin space 20 to the power of ring operator open parentheses 4 plus sec space 20 to the power of ring operator close parentheses$ =

In the figure $P Q greater than P R comma$ QS and RS are the bisectors of $stack ∣ Q with _ below$ and $stack R with _ below$ respectively then

In the figure $P Q greater than P R comma$ QS and RS are the bisectors of $stack ∣ Q with _ below$ and $stack R with _ below$ respectively then

If AB=AC and $A C D equals 115 to the power of ring operator end exponent$ then $stack B with _ below A C equals$

If AB=AC and $A C D equals 115 to the power of ring operator end exponent$ then $stack B with _ below A C equals$

In the figure $A equals 30 to the power of ring operator end exponent & E B equals 50 to the power of ring operator end exponent$ then ACD=

In the figure $A equals 30 to the power of ring operator end exponent & E B equals 50 to the power of ring operator end exponent$ then ACD=

$open parentheses sin to the power of 8 space 75 to the power of ring operator minus cos to the power of 8 space 75 to the power of ring operator close parentheses$ =

$open parentheses sin to the power of 8 space 75 to the power of ring operator minus cos to the power of 8 space 75 to the power of ring operator close parentheses$ =

The minimum value $m space equals bold space 2 cos space theta plus fraction numerator 1 over denominator sin invisible function application theta end fraction plus square root of 2 tan space theta space text in end text open parentheses 0 comma pi over 2 close parentheses$ is

The minimum value $m space equals bold space 2 cos space theta plus fraction numerator 1 over denominator sin invisible function application theta end fraction plus square root of 2 tan space theta space text in end text open parentheses 0 comma pi over 2 close parentheses$ is

In the adjoining figure, $stack left floor a with _ below equals left floor b$ and $left floor c equals left floor d comma$ then $left floor b plus left floor c equals$

In the adjoining figure, $stack left floor a with _ below equals left floor b$ and $left floor c equals left floor d comma$ then $left floor b plus left floor c equals$

In the figure, $a equals 4 x to the power of ring operator end exponent$ and $b equals left parenthesis 2.5 x plus 24 right parenthesis to the power of ring operator end exponent comma$ then the value of x =

In the figure, $a equals 4 x to the power of ring operator end exponent$ and $b equals left parenthesis 2.5 x plus 24 right parenthesis to the power of ring operator end exponent comma$ then the value of x =

In the given figure, if $P Q R equals 70 to the power of ring operator end exponent$ then $∣ A B C equals$

In the given figure, if $P Q R equals 70 to the power of ring operator end exponent$ then $∣ A B C equals$