Question

If cosθ = 2p / p²+1,(p ≠ 0)then sinθ is equal to:
1. 2p / p² +1
2. p²+1/ p²-1
3. 2p / p²-1
4. p²-1/2p

Answer 1

GIVEN :

If $cos\theta = \frac{2p}{ p^2+1}$, $(p\neq 0)$

TO FIND :

The value of $sin\theta$

SOLUTION :

Given that $cos\theta = \frac{2p}{ p^2+1}$, $(p\neq 0)$

The trignometric formula is given by :

$sin^2x=1-cos^2x$

Substitute the value of $cos\theta = \frac{2p}{p^2+1}$, $(p\neq 0)$ in the above formula we get

$sin^2\theta =1-(\frac{2p}{p^2+1})^2$, $(p\neq 0)$

$=1-\frac{(2p)^2}{(p^2+1)^2}$

By using the property of exponents of power rule :

$(ab)^m=a^mb^m$

By using the  Algebraic identity :

$(a+b)^2=a^2+2ab+b^2$

$=1-\frac{2^2(p)^2}{(p^2)^2+2(p^2)(1)+(1)^2}$

$=1-\frac{4p^2}{p^4+2p^2+1}$

$=\frac{p^4+2p^2+1-4p^2}{p^4+2p^2+1}$

$=\frac{p^4-2p^2+1}{p^4+2p^2+1}$

By using the  Algebraic identities :

$(a+b)^2=a^2+2ab+b^2$

$(a-b)^2=a^2-2ab+b^2$

$sin^2\theta=\frac{(p^2-1)^2}{(p^2+1)^2}$ , $(p\neq 0)$

Taking square root on both sides we get,

∴ $sin\theta=\frac{p^2-1}{p^2+1}$ , $(p\neq 0)$

∴ option 2) $\frac{p^2-1}{p^2+1}$ , $(p\neq 0)$ is correct.

∴ the value of $sin\theta$ is $\frac{p^2-1}{p^2+1}$ , $(p\neq 0)$