Question

√1-sin/1+sin=sec-tan

Answer 1

Heya!!✌

A good question from trigonometry,you have asked. Here is your answer.Please consider A as theta.
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√1-sinA\√1+sinA
=> (√1-sinA×√1-sinA)\(√1+sinA ×√1-sinA)
=> √(1-sinA)²\ √1²-sin²A
=> 1-sinA\√cos²A [ 1-sin²A= cos²A]
=> 1-sinA\cosA
=> (1\cosA) - (sinA\cosA)
=> secA - tanA
[Showed]
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Friend, hope you can do this type of math after understanding this.

Answer 2

To prove that: $\sqrt{\dfrac{1-\sin \theta}{1+\sin \theta} } =\sec \theta-\tan \theta.$

Solution:

L.H.S. = $\sqrt{\dfrac{1-\sin \theta}{1+\sin \theta} }$

To rationalizing denominator, we get

= $\sqrt{\dfrac{1-\sin \theta}{1+\sin \theta} }$ × $\sqrt{\dfrac{1-\sin \theta}{1-\sin \theta} }$

Using the algebraic identity:

(a + b)(a - b) = $a^{2}$ - $b^{2}$

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

= $\dfrac{1-\sin \theta}{\sqrt{1-\sin^2 \theta}}$

Using the trigonometric identity:

$\sin^2 A$ + $\cos^2 A$ = 1

⇒ $\cos^2 A$ = 1 - $\sin^2 A$

= $\dfrac{1-\sin \theta}{\sqrt{\cos^2 \theta} }$

= $\dfrac{1-\sin \theta}{\cos \theta}$

= $\dfrac{1}{\cos \theta}-\dfrac{\sin \theta}{\cos \theta}$

= $\sec \theta-\tan \theta$

= R.H.S., proved.

Thus, $\sqrt{\dfrac{1-\sin \theta}{1+\sin \theta} } =\sec \theta-\tan \theta$, proved.