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Maths-

General

Easy

Question

If $sin space x plus cosec space x plus tan space y plus cot space y equals 4$ where x and y $element of open parentheses 0 comma pi over 2 close parentheses tan space y over 2$ is a root of the equation

$α^{2} + 2 α + 1 = 0$
$α^{2} + 2 α - 1 = 0$
$2 α^{2} - 2 α - 1 = 0$
$α^{2} - α - 1 = 0$

The correct answer is: $alpha squared plus 2 alpha minus 1 equals 0$

$U sin g space A. M. greater or equal than G. M. s i n x plus cos e c x greater or equal than 2 a n d space tan x plus c o t x greater or equal than 2 S o comma space sin x plus cos e c x plus tan x plus c o t x greater or equal than 4 T h i s space i s space p o s s i b l e space o n l y space w h e n space x equals pi over 2 a n d y equals pi over 4 a n d space t an y equals 1 rightwards double arrow tan y equals 1 fraction numerator 2 tan begin display style y over 2 end style over denominator 1 minus tan squared begin display style y over 2 end style end fraction equals 1 1 minus tan squared y over 2 equals 2 tan y over 2 tan squared y over 2 plus 2 tan y over 2 minus 1 equals 0 i. e. space alpha squared plus 2 alpha minus 1 equals 0$

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If $fraction numerator tan space left parenthesis alpha plus beta minus gamma right parenthesis over denominator tan space left parenthesis alpha minus beta plus gamma right parenthesis end fraction equals fraction numerator tan begin display style space end style gamma over denominator tan begin display style space end style beta end fraction left parenthesis beta not equal to gamma right parenthesis$ then $sin space 2 alpha plus sin space 2 beta plus sin space 2 gamma$ =

If $fraction numerator tan space left parenthesis alpha plus beta minus gamma right parenthesis over denominator tan space left parenthesis alpha minus beta plus gamma right parenthesis end fraction equals fraction numerator tan begin display style space end style gamma over denominator tan begin display style space end style beta end fraction left parenthesis beta not equal to gamma right parenthesis$ then $sin space 2 alpha plus sin space 2 beta plus sin space 2 gamma$ =

If $fraction numerator sin cubed space theta minus cos cubed space theta over denominator sin space theta minus cos space theta end fraction minus fraction numerator cos begin display style space end style theta over denominator square root of open parentheses 1 plus cot squared space theta close parentheses end root end fraction$ -2 tan $theta cot space theta equals negative 1 comma theta element of left square bracket 0 comma 2 pi right square bracket$ then

If $fraction numerator sin cubed space theta minus cos cubed space theta over denominator sin space theta minus cos space theta end fraction minus fraction numerator cos begin display style space end style theta over denominator square root of open parentheses 1 plus cot squared space theta close parentheses end root end fraction$ -2 tan $theta cot space theta equals negative 1 comma theta element of left square bracket 0 comma 2 pi right square bracket$ then

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The reaction (CH₃)₃ C - Br → (CH₃)₃ $stack C with plus on top plus B r to the power of minus end exponent$ is an example of

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$Na subscript 2 straight B subscript 4 straight O subscript 7 times 10 straight H subscript 2 straight O not stretchy rightwards arrow with straight capital delta on top NaBO subscript 2 plus left parenthesis A right parenthesis plus straight H subscript 2 straight O left parenthesis A right parenthesis plus MnO not stretchy rightwards arrow with straight capital delta on top left parenthesis B right parenthesis$ (A) and (B) are

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