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If secθ + tanθ = p, show that (p^2−1)/(p^2+1)=sinθ - Mathematics

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Sum

If secθ + tanθ = p, show that `(p^{2}-1)/(p^{2}+1)=\sin \theta`

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Solution 1

We have,

`=(\sec ^{2}\theta +\tan ^{2}\theta +2\sec \theta \tan\theta -1)/(\sec ^{2}\theta +\tan^{2}\theta +2\sec \theta \tan\theta +1)`

`=\frac{(\sec ^{2}\theta -1)+\tan ^{2}\theta +2\sec \theta \tan\theta }{\sec ^{2}\theta +2\sec \theta \tan \theta +(1+\tan^{2}\theta )`

`=(\tan ^{2}\theta +\tan ^{2}\theta +2\sec \theta \tan\theta )/(\sec ^{2}\theta +2\sec \theta \tan \theta +\sec^{2}\theta )`

`=\frac{2\tan ^{2}\theta +2\tan \theta \sec \theta }{2\sec^{2}\theta +2\sec \theta \tan \theta }`

`=\frac{2\tan \theta (\tan \theta +\sec \theta )}{2\sec \theta (\sec\theta +\tan \theta )}`

`=\frac{\tan \theta }{\sec \theta }=\frac{\sin \theta }{\cos \theta \sec\theta }`

= sinθ = RHS

Solution 2

Sec θ + tan θ = P

⇒ `1/cos θ + sin θ /cos θ  = P`

⇒ `(1 + sin θ)/cos θ = P`

⇒ `(1 + sin θ)^2/cos^2 θ = P^2`,      ....(Squaring both sides)

⇒ `(1 + sin^2 θ + 2 sin θ)/cos^2 θ = p^2`

⇒ `(1 + sin^2 θ + 2 sin θ  + cos^2 θ)/(1 + sin^2 θ + 2 sin θ  - cos^2 θ) = (p^2 + 1)/(p^2 - 1)`   ....(Applying componendo and dividendo]

⇒ `(1 + 1 + 2 sin θ)/(sin^2 θ + sin^2 θ + 2 sin θ) = (p^2 + 1)/(p^2 - 1)`

⇒ `(2( 1 + sin θ))/(2 sin θ( 1 + sin θ)) = (p^2 + 1)/(p^2 - 1)`

⇒ `1/sin θ = (p^2 + 1)/(p^2 - 1)`

Taking reciprocals, we get,

⇒ sin θ = `(p^2 - 1)/(p^2 + 1)`

Hence proved.

Concept: Trigonometric Identities
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APPEARS IN

ICSE Class 10 Mathematics
Chapter 18 Trigonometry
Exercise | Q 43
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