Question

sin^2 a +sin^2 a ×tan^2 a =tan^2 a prove that
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Answer 1

Correct Question

$\sin {}^{2} a \tan {}^{2} a = \tan {}^{2} a - \sin {}^{2} a$

Given :-

$\sin {}^{2} a \tan {}^{2} a = \tan {}^{2} a - \sin {}^{2} a$

To Prove :-

$\sin {}^{2} a \tan {}^{2} a = \tan {}^{2} a - \sin {}^{2} a$

Proof :-

Let us first find the value of right hand side (RHS) that is tan²a - sin²a as shown below :

$\tan {}^{2} a - \sin {}^{2} a$

$= \frac{ \sin {}^{2} a }{ \cos{}^{2}a } - \sin {}^{2} a$

$(∵ \tan \: x = \frac{ \sin \: x }{ \cos \: x } )$

$= \frac{ \sin {}^{2}a - \sin {a}^{2} \cos {}^{2} a }{\cos {}^{2} a} \\ = \sin {}^{2}a(\frac{1 - \cos {}^{2} a}{\cos {}^{2} a} ) \: \: \: \: \: \\ = \frac{\sin {}^{2}a}{\cos {}^{2} a} (1 - \cos {}^{2} a) \: \: \: \: \\ = \tan {}^{2} a \sin {}^{2} a = RHS$

$(∵ \tan \: x = \frac{ \sin \: x }{ \cos \: x} \: ,1 - \cos {}^{2} x = \sin {}^{2} x)$

Since RHS = LHS

Hence,

$\sin {}^{2} a \tan {}^{2} a = \tan {}^{2} a - \sin {}^{2} a$

Hence Proved !