Question

Plz help me with at the earliest.....I will mark the ans as brainliest if get a good ans

Answer 1

GIVEN:

• $\sf{\dfrac{1- sin \theta }{cos \theta} + \dfrac{cos \theta }{1- sin \theta} = 2 sec \theta}$

TO FIND:

• Evaluate

SOLUTION:

Take L.H.S.

$\rm{\longrightarrow \dfrac{1- sin \theta }{cos \theta} + \dfrac{cos \theta }{1- sin \theta} }$

$\rm{\longrightarrow \dfrac{(1-sin \theta)^2 + cos^2 \theta}{cos \theta (1- sin \theta)} }$

Using Identity

• (a–b)² = a² –2ab + b²

$\rm{\longrightarrow \dfrac{1 - 2sin \theta + sin^2 \theta + cos^2 \theta }{cos \theta (1- sin \theta)}}$

Using Trigonometric Identity

• sin²θ + cos²θ = 1

$\rm{\longrightarrow \dfrac{1+ 1 - 2 sin \theta}{cos \theta (1- sin \theta)}}$

$\rm{\longrightarrow \dfrac{2 - 2 sin \theta}{cos \theta (1- sin \theta)} }$

$\rm{\longrightarrow \dfrac{2 \cancel{(1- sin \theta)}}{cos \theta \cancel{(1- sin \theta)}}}$

$\rm{\longrightarrow \dfrac{2}{cos \theta}}$

We know that, 1/cosθ = secθ

$\bf{\longrightarrow 2 sec \theta = R.H.S.}$

$\Large{\underline{\underline{\bf{Hence \: Proved \text{!}}}}}$

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Answer 2

LHS = cos theta / 1-sin theta + 1-sin theta / cos theta

= cos theta * cos theta + 1-sin theta * 1-sin theta / cos theta * 1-sin theta

= cos sq theta + 1+ sin sq theta -2sin theta / cos theta * 1-sin theta

= 2 -2sin theta / cos theta * 1-sin theta

=2(1-sin theta )/ cos theta * 1-sin theta

=2/cos theta

= 2 * 1/cos theta

=2 sec theta

= RHS