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प्रश्न
Prove the following identity :
`(sec^2θ - sin^2θ)/tan^2θ = cosec^2θ - cos^2θ`
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उत्तर
LHS = `(sec^2θ - sin^2θ)/tan^2θ`
= `(1/cos^2θ - sin^2θ)/(sin^2θ/cos^2θ)`
= `(1 - sin^2θcos^2θ)/((cos^2θ)/(sin^2θ/cos^2θ))`
= `(1 - sin^2θcos^2θ)/sin^2θ`
= `1/sin^2θ - (sin^2θcos^2θ)/(sin^2θ)`
= `cosec^2θ - cos^2θ`
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संबंधित प्रश्न
Prove that ` \frac{\sin \theta -\cos \theta +1}{\sin\theta +\cos \theta -1}=\frac{1}{\sec \theta -\tan \theta }` using the identity sec2 θ = 1 + tan2 θ.
Prove the following trigonometric identities.
`[tan θ + 1/cos θ]^2 + [tan θ - 1/cos θ]^2 = 2((1 + sin^2 θ)/(1 - sin^2 θ))`
`(1+ tan^2 theta)/(1+ tan^2 theta)= (cos^2 theta - sin^2 theta)`
Prove that:
`"tan A"/(1 + "tan"^2 "A")^2 + "Cot A"/(1 + "Cot"^2 "A")^2 = "sin A cos A"`.
Prove the following identity :
`(secθ - tanθ)^2 = (1 - sinθ)/(1 + sinθ)`
Prove the following identity :
`(cot^2θ(secθ - 1))/((1 + sinθ)) = sec^2θ((1-sinθ)/(1 + secθ))`
Evaluate:
`(tan 65^circ)/(cot 25^circ)`
tan θ cosec2 θ – tan θ is equal to
a cot θ + b cosec θ = p and b cot θ + a cosec θ = q then p2 – q2 is equal to
If sec θ = `25/7`, find the value of tan θ.
Solution:
1 + tan2 θ = sec2 θ
∴ 1 + tan2 θ = `(25/7)^square`
∴ tan2 θ = `625/49 - square`
= `(625 - 49)/49`
= `square/49`
∴ tan θ = `square/7` ........(by taking square roots)
