Question

Assertion :If ${\tan }^{2}A={\tan }^{2}B,{\cos }^{2}A={\cos }^{2}B{\sin }^{2}A={\sin }^{2}B$, then $A=n\pi \pm B,nϵI$ Reason: $\tan A=\tan B$, $\cos A=\cos B,\sin A=\sin B$, then $A=n\pi \pm B,nϵI$

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Assertion false but Reason is true

Answer 1

The correct option is C Assertion is correct but Reason is incorrect

${\sin }^{2}A={\sin }^{2}B$ .....( i )
$\Rightarrow 1-{\sin }^{2}A=1-{\sin }^{2}B$
$\Rightarrow {\cos }^{2}A={\cos }^{2}B$ ..............( ii )
$\Rightarrow {\tan }^{2}A={\tan }^{2}B$ (On diving (i) by (ii ))
$\Rightarrow {\tan }^{2}A-{\tan }^{2}B=0$
$\Rightarrow \tan (A+B)\tan (A-B)=0$
$\Rightarrow A=n\pi \pm B(n\in I)$
Again $\sin A=\sin B$
$A=n\pi +(-1{)}^{n}B$
($\therefore A=n\pi \pm B$ accordingly $n$ is even or odd integer)
And $\cos A=\cos B$
$\Rightarrow A=n\pi \pm B(n\in I)$
Also $\tan A=\tan B$
$\tan A-\tan B=0$
$\tan (A-B)=0$
$A-B=n\pi$
$\therefore A=n\pi +B(n\in I)$
$\therefore$ Assertion (A) is true but Reason (R) is false,