Question

( sec theta - tan theta ) whole square ( 1 + sin theta ) = 1 - sin theta ​

Answer 1

Answer:

Instead of theta, I have used alpha. Replace the alpha with theta.

Step-by-step explanation

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

Answer 2

Answer:

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