Question

Please answer step by step attachment

Answer 1

Question

Prove that

• 1/( sec θ - tan θ ) = (1 + sin θ)/ cos θ

Prove

First take L.H.S.

==> 1/( sec θ - tan θ )

Multiply by ( sec θ + tan θ) in Numerator & denominator

==> (sec θ + tan θ)/(sec θ + tan θ)(sec θ - tan θ)

==> (sec θ + tan θ)/(sec² θ - tan² θ)

Using Formula

• (sec² θ - tan² θ) = 1

==> (sec θ + tan θ)/1

==> (sec θ + tan θ)

Now, take R.H.S.

==> (1 + sin θ)/ cos θ

==> 1/cos θ + sin θ/ cos θ

[ ★ 1/cos θ = sec θ , sin θ/ cos θ = tan θ ]

==>( sec θ + tan θ)

Here,

L.H.S. = R.H.S.

Hence, Proved

_________________

Answer 2

Answer:

$\frac{1}{sec \theta - tan \theta} = \frac{1 + sin \theta}{cos \theta} \\ \\ \\ = > \frac{1}{ \frac{1}{cos \theta} - \frac{sin \theta}{cos \theta} } \\ \\ \\ = > \frac{cos \theta}{1 - sin \theta} \times \frac{1 + sin \theta}{1 + sin \theta} \\ \\ \\ = > \frac{cos \theta(1 + sin \theta)}{(1 - sin \theta)(1 + sin \theta)} \\ \\ \\ you \: know... \\ (x + y)(x - y) = {x}^{2} - {y}^{2} \\ \\ \\ so... \\ \\ \\ = = > > \frac{cos \theta(1 + sin \theta)}{1 - {sin}^{2} \theta} \\ \\ \\ \\ = = > > \frac{cos \theta(1 + sin \theta)}{cos ^{2} \theta} \\ \\ \\ = = > > \frac{(1 + sin \theta)}{cos \theta} \: \: proved....$

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