Question

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

Answer 1

Hey !!!

( 1 + tan²¢) ( 1 + sin¢ ) ( 1 - sin¢)

( 1 + tan²¢ ) ( 1 - sin²¢)

we know that , (1 + tan²¢ ) = sec²¢
and ( 1 - sin²¢ ) = cos²¢

since,

sec²¢ × cos²¢

but sec²¢ = 1/cos²¢

So , 1/cos²¢ × cos²¢ = 1

Rhs = LHS prooved ♻

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Hope it helps you !!

@Rajukumar111 :-D

Answer 2

$\bf \: Taking \: LHS \: \: , \\ \\ \bf{(1 + \tan {}^{2} ( \alpha ) )(1 + \sin( \alpha )) (1 - \sin( \alpha ))} \\ \\ \bf \: {(1 + \tan {}^{2} ( \alpha ) )(1 - \sin {}^{2} ( \alpha ) )} \\ \\ \bf \sec {}^{2} ( \alpha ) \: \times \: \cos {}^{2} ( \alpha ) = \frac{1}{ \cos {}^{2} ( \alpha ) } \times \cos {}^{2} ( \alpha ) \\ \\ = \: \: \: 1 \: \: = RHS$