Question

pls help and no spam.... ​

Answer 1

Answer:

[9/18, 2:24 PM] Zenifer: Mam maths ka Monday ko test hoga

[9/18, 2:24 PM] Mam Yusra Class Teacher: G

[9/18, 2:24 PM] Zenifer: Mam kon kon Şi exercises ka

[9/18, 2:25 PM] Mam Yusra Class Teacher: 6.1 to 6.4

[9/18, 2:25 PM] Zenifer: Okay mam

Answer 2

GIVEN★

RHS

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

=LHS

$\cos {}^{2} ( \alpha ) - \sin {}^{2} ( \alpha )$

VERIFIED★

LHS=RHS

EVALUTION★

LHS

• $\frac{1 - \tan {}^{2} ( \alpha ) }{1 + \tan {}^{2} ( \alpha ) }$
• we know that

• $1 + \tan {}^{2} ( \alpha ) = \sec {}^{2} ( \alpha )$
• $\frac{1 - \tan {}^{2} ( \alpha ) }{ \sec {}^{2} ( \alpha ) } \\ \\ = \frac{1}{ \sec {}^{2} ( \alpha ) } - \frac{ \tan {}^{2} ( \alpha ) }{ \sec {}^{2} ( \alpha ) }$
• we know that

• $\frac{1}{ \sec( \alpha ) } = \cos( \alpha ) \\ \\ \tan( \alpha ) = \frac{ \sin( \alpha ) }{ \cos( \alpha ) }$
• $= \cos {}^{2} ( \alpha ) - \frac{ \frac{ \sin {}^{2} ( \alpha ) }{ \cos {}^{2} ( \alpha ) } }{ \frac{1}{ \cos {}^{2} ( \alpha ) } } \\ \\ = \cos {}^{2} ( \alpha ) - \frac{ \sin {}^{2} ( \alpha ) \cos {}^{2} ( \alpha ) }{ \cos {}^{2} ( \alpha ) } \\ \\ = \cos {}^{2} ( \alpha ) - \sin {}^{2} ( \alpha )$

RHS =LHS (hence proved)

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I hope it helps