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

Sum

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

#### 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.

Is there an error in this question or solution?
Chapter 18: Trigonometry - Exercise 2

#### APPEARS IN

ICSE Class 10 Mathematics
Chapter 18 Trigonometry
Exercise 2 | Q 43

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