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].
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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)
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(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