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Maths-

General

Easy

Question

$Tan space open parentheses pi over 4 plus A over 2 close parentheses plus Tan space open parentheses pi over 4 minus A over 2 close parentheses equals fraction numerator 4 over denominator square root of 3 end fraction$ ; then the general solution of A=

$2 n π \pm \frac{π}{6}, \forall n \in Z$
$n π + \frac{π}{4}, \forall n \in Z$
$2 n π \pm \frac{π}{4}, \forall n \in Z$
$n π, \forall n \in Z$

Hint:

A general solution is one which involves the integer ‘n’ and gives all solutions of a trigonometric equation.

The correct answer is: $2 n pi plus-or-minus pi over 6 comma straight for all n element of Z$

We have, tan(π/4 + A/2) + tan (π/4 - A/2) = 4/√3
=> (1+ tanA/2)/(1-tanA/2) + (1-tanA/2)/(1+tanA/2) = 4/√3
=> {(1 + tan²A/2 + 2tanA/2) + (1 + tan²A/2 - 2tanA/2)}/ (1- tan²A/2) = 4/√3
=> {2(1+ tan²A/2)}/(1 - tan²A/2) = 4/√3
=> (1-tan²A/2)/(1+tan²A/2) = √3/2
=> cos 2(A/2) = cos π/6
=> cosA = cos π/6
=> A = 2nπ ± π/6.
Hence, the correct option is C.

tan (π/4 + A) = (1+tanA)/(1-tanA).

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If $11 sin squared space x plus 7 Cos squared space x equals 8$ then x = - ....

Here, we have to find the value of x.
11 sin²x + 7 cos²x = 8 .
Or, 11 sin²x + 7 ( 1 - sin²x) = 8 .
Or, 11 sin²x + 7 - 7sin²x = 8 .
Or, 11sin²x - 7 sin²x = 8 - 7 .
Or, 4sin²x = 1
Or, sin²x = 1/4
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