Question

root 1- sec theta/1+ sin theta= sec theta- tan theta​

Answer 1

Step-by-step explanation:

Correct Question:-

• $\rm \: Prove \: that : \sqrt{ \dfrac{1 - \sin\theta }{1 + \sin\theta} } = \sec\theta - \tan\theta$

$\bf{\underline{\underline\red{To\:Prove}}}$

• $\sqrt{ \dfrac{1 - \sin\theta }{1 + \sin\theta} } = \sec\theta - \tan\theta$

$\bf{\underline{\underline\green{Proof:-}}}$

Let us prove by simplifying LHS and RHS seperately

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

By rationalising:-

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

$\rm = \sqrt{ \dfrac{(1 - \sin\theta) (1 - \sin\theta)}{(1 + \sin\theta)(1 - \sin\theta)} }$

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

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

$\rm = \sqrt{ \dfrac{(1 - \sin\theta)^{2} }{ cos ^{2} \theta} } \: \: \: \: \: \: \: \: \bigg( \because1 - {sin}^{2} \theta = {cos}^{2} \theta \bigg)$

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

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

$\rm = \dfrac{1}{cos \theta} - \dfrac{sin \theta}{cos \theta} \:\:\: \:\: \: \bigg(\because\dfrac{1}{cos\theta} = sec\theta\: , \: \dfrac{sin\theta}{cos\theta} = tan\theta\bigg)$

$\rm = sec \theta - tan \theta = RHS$

LHS = RHS

Hence Proved

Answer 2

✔ Explanation :-

✔ Given :-

( Here, Theta = A)

• √1-SinA/1 + SinA = SecA - TanA

✔ Solution :-

=> LHS :-

=> √1 - SinA / 1 + SinA

Rationalize it,

=> √1 - SinA / 1 + SinA × 1 - SinA / 1 - SinA

=> √(1 - SinA).(1 - SinA) / ( 1 + SinA). (1 - SinA)

=> √(1 - SinA)² / (1)² - (SinA)²

=> √(1 - SinA)² / Cos²A

=> 1 - SinA / CosA

=> 1/CosA - SinA/CosA

=> SecA - TanA

=> RHS.

✔ Formulae Used :-

• Sin²A + Cos²A = 1