Question

To prove:
$(1 - sin \: a \:) \div (1+ sin \: a )\: = (sec \: a\: - tan \: a) {}^{2}$
pls give step by step answer.
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Answer 1

$\rm{(1-\sin a) \div(1+\sin a)=(\sec a-\tan a)^{2}}$

$\longrightarrow\rm{(\sec a-\tan a)^{2}=(1-\sin a) \div(1+\sin a)}$

$\longrightarrow\rm{(1/\cos a-\sin a/\cos a)^{2}=(1-\sin a) \div(1+\sin a)}$

$\longrightarrow\rm{[(1-\sin a)/\cos a]^{2}=(1-\sin a) \div(1+\sin a)}$

$\longrightarrow\rm{(1-\sin a)^{2}/\cos^{2} a=(1-\sin a) \div(1+\sin a)}$

$\longrightarrow\rm{(1-\sin a)^{2}/(1-\sin^{2} a)=(1-\sin a) \div(1+\sin a)}$

$\longrightarrow\rm{(1-\sin a)^{2}/[(1-\sin a)(1+\sin a)]=(1-\sin a) \div(1+\sin a)}$

$\longrightarrow\rm{[(1-\sin a)(1-\sin a)]/[(1-\sin a)(1+\sin a)]=(1-\sin a) \div(1+\sin a)}$

$\boxed{\longrightarrow\rm{(1-\sin a)/(1+\sin a)=(1-\sin a) \div(1+\sin a)}}$

Answer 2

Step-by-step explanation:

To prove:-

$\sf{\dfrac{1-sin \theta}{ 1 + sin \theta} = (sec \theta - tan \theta)}$

Solution :-

$\sf{\dfrac{1-sin \theta}{ 1 + sin \theta}}$

Rationalise it with ( 1 - sin theta)

$\sf{\dfrac{1-sin \theta \times (1 - sin \theta)}{ 1 + sin \theta \times (1 - sin \theta)}}$

$\sf{\dfrac{(1-sin \theta)^2}{ 1 - (sin \theta)^2}}$

Used a²-b² identity in denominator.

$\sf{\dfrac{(1-sin \theta)^2}{ 1 - sin^2 \theta}}$

$\sf{\dfrac{1 + sin^2 \theta - 2sin \theta }{ 1 - sin^2 \theta}}$

$\sf{\dfrac{1 + sin^2 \theta - 2sin \theta }{ cos^2 \theta}}$

Divide it accordingly as :-

$\sf{\dfrac{1 + sin^2 \theta }{cos^2 \theta} \dfrac{- 2sin \theta}{ cos^2 \theta}}$

$\sf{\dfrac{1}{cos^2 \theta} + \dfrac{sin^2 \theta}{cos^2 \theta} - \dfrac{2}{cos \theta} \times \dfrac{sin \theta}{cos \theta}}$

We know that:-

$\sf{\dfrac{sin \theta}{cos \theta} = tan \theta}$

$\sf{\dfrac{1}{cos \theta} = sec \theta}$

Placing the values, we get :-

$\sf{sec^2 \theta + tan^2 \theta - 2 sec \theta tan \theta}$

Simplifying, we get :-

$\sf{(sec \theta - tan \theta)^2}$

Hence Proved!