The value of tan^(-1)(cos15π)/4-1sin(π)/4 is equal to

Correct answer: (B) -(π)/8. Given, tan^(-1)(cos15π)/4-1sin(π)/4 =tan^(-1)(cos4π-π4-1)/(sinπ4) =tan^(-1)(1√2-1)/(1√2) =tan^(-1)1-√2 =-tan^(-1)√2-1 =-(π)/8

Q48 / 90 · JEE Main 2022 (25 Jun Shift 2) · Mathematics

The value of $\tan^{- 1} [\frac{\cos (\frac{15 π}{4}) - 1}{\sin (\frac{π}{4})}]$ is equal to

A
$- \frac{π}{4}$
Correct answer
B
$- \frac{π}{8}$
C
$- \frac{5 π}{12}$
D
$- \frac{4 π}{9}$

Solution

Given,

$\tan^{- 1} (\frac{\cos (\frac{15 π}{4}) - 1}{\sin \frac{π}{4}})$

$= \tan^{- 1} (\frac{\cos (4 π - \frac{π}{4}) - 1}{\sin \frac{π}{4}})$

$= \tan^{- 1} (\frac{\frac{1}{\sqrt{2}} - 1}{\frac{1}{\sqrt{2}}})$

$= \tan^{- 1} (1 - \sqrt{2}) = - \tan^{- 1} (\sqrt{2} - 1)$

$= - \frac{π}{8}$

Hence, the correct answer is option: 'B'.

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The value of tan-1cos15π4-1sinπ4 is equal to

Solution

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The value of $\tan^{- 1} [\frac{\cos (\frac{15 π}{4}) - 1}{\sin (\frac{π}{4})}]$ is equal to