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Simplify: 1(secθ−tanθ) \frac{1}{(sec θ - tan θ)}(secθ−tanθ)1​  = ? ​
Question

Simplify: 1(secθtanθ) \frac{1}{(sec θ - tan θ)}  = ? ​

A.

sec² θ - tan² θ

B.

cos θ + sin θ

C.

sec θ + tan θ

D.

cosec θ - cot θ

Correct option is C

Given:

1secθtanθ\frac{1}{\sec \theta - \tan \theta}​​

Formula Used:

Trigonometric identities:
secθ=1cosθ tanθ=sinθcosθ\sec \theta = \frac{1}{\cos \theta}\\\ \\\tan \theta = \frac{\sin \theta}{\cos \theta}​​

Solution:

1secθtanθ =11cosθsinθcosθ =11sinθcosθ =cosθ1sinθ \frac{1}{\sec \theta - \tan \theta}\\\ \\= \frac{1}{\frac{1}{\cos \theta} - \frac{\sin \theta}{\cos \theta}}\\\ \\= \frac{1}{\frac{1 - \sin \theta}{\cos \theta}}\\\ \\= \frac{\cos \theta}{1 - \sin \theta}\\\ \\

=cosθ1sinθ×1+sinθ1+sinθ = \frac{\cos \theta}{1 - \sin \theta}\times\frac{1 +\sin \theta}{1 +\sin \theta}\\\ \\

=(cosθ)(1+sinθ)1sin2θ = \frac{(\cos \theta)(1 +\sin \theta)}{1 -\sin^2 \theta}\\\ \\

=(cosθ)(1+sinθ)cos2θ = \frac{(\cos \theta)(1 +\sin \theta)}{cos^2 \theta}\\\ \\

=(1+sinθ)cosθ = \frac{(1 +\sin \theta)}{cos \theta}\\\ \\ 

=secθ+tanθ = sec \theta +{tan \theta}\\\ \\​​

​​​

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