Question

If $\tan \theta =\frac{x^2+1}{x^2-1}$, then

• A. $\sin \theta =\frac{x^2+1}{\sqrt{2(x^4+1)}}$
• B. $\sin \theta =\frac{x^2-1}{\sqrt{2(x^4+1)}}$
• C. $\cos \theta =\frac{x^2+1}{\sqrt{2(x^4+1)}}$
• D. $\cos \theta =\frac{x^2-1}{\sqrt{2(x^4+1)}}$

Answer 1

Answer: (A), (D)

The correct options are
A $\sin \theta =\frac{x^2+1}{\sqrt{2(x^4+1)}}$
D $\cos \theta =\frac{x^2-1}{\sqrt{2(x^4+1)}}$

Let $\tan \theta =\frac{x^2+1}{x^2-1}=\frac{P}{B}$
Then $P=x^2+1$ and $B=x^2-1$
Using Pythagoras theorem
$H=\sqrt{P^2+B^2}=\sqrt{{(x^2+1)}^2+{(x^2-1)}^2}=\sqrt{x^4+1+2x^2+x^4+1-2x^2}=\sqrt{2(x^4+1)}$
We get
$\sin \theta =\frac{P}{H}=\frac{x^2+1}{\sqrt{2(x^4+1)}}$
And
$\cos \theta =\frac{B}{H}=\frac{x^2-1}{\sqrt{2(x^4+1)}}$