Question

Prove that :- (secA-tanA)^2(1+sin A) = 1-sin A​

Answer 1

Answer:

LHS = RHS

True

Step-by-step explanation:

sec A = 1/cos A

tan A= sin A / CosA

Well we can write LHS :

( 1/cos A - sin A / Cos A)² ( 1+ Sin A)

combine the fraction

(1 - Sin A/cos A)² (1+sin A)

(1-sin A)²/ Cos² A ( 1 + Sin A)

Mulitply fractions

(1-sin A)²(1+ Sin A) / cos² A

By pythagoras identity:

cos ² A + Sin² x = 1

Cos² A = 1 - Sin ² A

We can write : 1 - sin² A instead of cos ² A

(1- sin A)²(1+sin A) / 1 - Sin² A

Factors of 1 - Sin²A : (1+sin A) ( 1 - SinA)

((1- Sin A)² (1+sin A))/ ((1+sinA)(1-sin A)

cancell common facto 1+sinA

(1-sin A)²/ ( 1- Sin A)

cancel comman factor : 1-sinA

I Showed that both sides are equal.

Mark me brainliest

Answer 2

Answer:

$\displaystyle{\boxed{\red{\sf\:(\:\sec\:A\:-\:\tan\:A\:)^2\:(\:1\:+\:\sin\:A\:)\:=\:1\:-\:\sin\:A}}}$

Step-by-step-explanation:

We have given a trigonometric equation.

We have to prove that equation.

The given trigonometric equation is

$\displaystyle{\sf\:(\:\sec\:A\:-\:\tan\:A\:)^2\:(\:1\:+\:\sin\:A\:)\:=\:1\:-\:\sin\:A}$

Now,

$\displaystyle{\sf\:(\:\sec\:A\:-\:\tan\:A\:)^2\:(\:1\:+\:\sin\:A\:)\:=\:1\:-\:\sin\:A}$

$\displaystyle{\sf\:LHS\:=\:(\:\sec\:A\:-\:\tan\:A\:)^2\:(\:1\:+\:\sin\:A\:)}$

$\displaystyle{\implies\sf\:LHS\:=\:\left(\:\dfrac{1}{(\:\sec\:A\:+\:\tan\:A\:)}\:\right)^2\:(\:1\:+\:\sin\:A\:)\:\:-\:-\:-\:[\:\because\:\sec^2\:A\:-\:\tan^2\:A\:=\:1\:]}$

$\displaystyle{\implies\sf\:LHS\:=\:\left(\:\dfrac{1}{\dfrac{1}{\cos\:A}\:+\:\dfrac{\sin\:A}{\cos\:A}}\:\right)^2\:(\:1\:+\:\sin\:A\:)\:-\:-\:-\:\left[\:\sec\:A\:=\:\dfrac{1}{\cos\:A}\:;\:\tan\:A\:=\:\dfrac{\sin\:A}{\cos\:A}\:\right]}$

$\displaystyle{\implies\sf\:LHS\:=\:\left(\:\dfrac{1}{\dfrac{1\:+\:\sin\:A}{\cos\:A}}\:\right)^2\:(\:1\:+\:\sin\:A\:)}$

$\displaystyle{\implies\sf\:LHS\:=\:\left(\:\dfrac{\cos\:A}{1\:+\:\sin\:A}\:\right)^2\:(\:1\:+\:\sin\:A\:)}$

$\displaystyle{\implies\sf\:LHS\:=\:\left(\:\dfrac{\cos^2\:A}{(\:1\:+\:\sin\:A\:)^2}\:\right)\:(\:1\:+\:\sin\:A\:)}$

$\displaystyle{\implies\sf\:LHS\:=\:\dfrac{1\:-\:\sin^2\:A}{(\:1\:+\:\sin\:A\:)\:(\:1\:+\:\sin\:A\:)}\:\times\:(\:1\:+\:\sin\:A\:)\:\:-\:-\:-\:[\:\because\:\sin^2\:A\:+\:\cos^2\:A\:=\:1\:]}$

$\displaystyle{\implies\sf\:LHS\:=\:\dfrac{\cancel{(\:1\:+\:\sin\:A\:)}\:(\:1\:-\:\sin\:A\:)}{\cancel{(\:1\:+\:\sin\:A\:)}\:(\:1\:+\:\sin\:A\:)}\:\times\:(\:1\:+\:\sin\:A\:)}$

$\displaystyle{\implies\sf\:LHS\:=\:\dfrac{(\:1\:-\:\sin\:A\:)}{\cancel{(\:1\:+\:\sin\:A\:)}}\:\times\:\cancel{(\:1\:+\:\sin\:A\:)}}$

$\displaystyle{\implies\sf\:LHS\:=\:(\:1\:-\:\sin\:A\:)}$

$\displaystyle{\implies\sf\:RHS\:=\:(\:1\:-\:\sin\:A\:)}$

$\displaystyle{\therefore\:\underline{\boxed{\red{\sf\:LHS\:=\:RHS\:}}}}$

Hence proved!