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प्रश्न
Prove the following trigonometric identities:
(i) (1 – sin2θ) sec2θ = 1
(ii) cos2θ (1 + tan2θ) = 1
योग
उत्तर
(i) We have,
`LHS = (1 – sin^2 θ) sec^2 θ`
`= cos^2 θ sec^2 θ [∵ 1 – sin^2 θ = cos^2 θ]`
`=cos ^{2}\theta ( {1}/cos ^{2}\theta)[ \because \ \ \sec \theta =\frac{1}{\cos \theta }]`
= 1 = RHS
(ii) We have,
`LHS = cos^2 θ (1 + tan^2 θ)`
`= cos^2 θ sec^2 θ [∵ 1 + tan^2 θ = sec^2 θ]`
`=\cos ^{2}\theta( 1/\cos^{2}\theta )[\because \ \ \sec \theta =\frac{1}{\cos \theta }]`
= 1 = RHS
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