Newbie

# Prove the following identities, where the angles involved are acute angles for which the expressions are defined.(iv) (1 + sec A)/sec A = sin2A/(1-cos A) [Hint : Simplify LHS and RHS separately]. Q.5(4)

• 0

Today i am solving trigonometry question  its very hard to solve please  help me to solve the ncert class 10 exercise 8.4 . Find the best way to solve question no.5(4) easily . Prove the following identities, where the angles involved are acute angles for which the expressions are defined.(iv) (1 + sec A)/sec A = sin2A/(1-cos A) [Hint : Simplify LHS and RHS separately].

Share

1. (1 + sec A)/sec A = sin2A/(1-cos A)

First find the simplified form of L.H.S

L.H.S. = (1 + sec A)/sec A

Since secant function is the inverse function of cos function and it is written as

= (1 + 1/cos A)/1/cos A

= (cos A + 1)/cos A/1/cos A

Therefore, (1 + sec A)/sec A = cos A + 1

R.H.S. = sin2A/(1-cos A)

We know that sin2A = (1 – cos2A), we get

= (1 – cos2A)/(1-cos A)

= (1-cos A)(1+cos A)/(1-cos A)

Therefore, sin2A/(1-cos A)= cos A + 1

L.H.S. = R.H.S.

Hence proved

• 0