Question

prove that tan a minus sin a by sin square A is equal to tan a by 1 + Cos A​

Answer 1

check the attachment

Answer 2

$\frac{\tan A-\sin A}{\sin^2 A}=\frac{\tan A}{1+\cos A}$ hence proved.

Step-by-step explanation:

To prove : $\frac{\tan A-\sin A}{\sin^2 A}=\frac{\tan A}{1+\cos A}$

Proof :

Taking LHS,

$LHS=\frac{\tan A-\sin A}{\sin^2 A}$

$LHS=\frac{\frac{\sin A}{\cos A}-\sin A}{1-\cos^2 A}$

$LHS=\frac{\frac{\sin A-\sin A\cos A}{\cos A}}{1-\cos^2 A}$

$LHS=\frac{\sin A(1-\cos A)}{\cos A(1+\cos A)(1-\cos A)}$

$LHS=\frac{\sin A}{\cos A(1+\cos A)}$

$LHS=\frac{\frac{\sin A}{\cos A}}{\frac{\cos A(1+\cos A)}{\cos A}}$

$LHS=\frac{\tan A}{1+\cos A}$

$LHS=RHS$

Hence proved.

#Learn more

Prove that tan square A minus sin square A is equals to tan square A into sin square A​

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