Question

If $sin\theta =\frac{a^2-b^2}{a^2+b^2}$, find the values of tan $\theta ,sec\theta$ and $cosec\theta .$

Answer 1

Now, cos $\theta =\sqrt{1-si{n}^2\theta }$

$=\sqrt{\frac{1-(a^2-b^2{)}^2}{(a^2+b^2{)}^2}}[\because sin\theta =\frac{a^2-b^2}{a^2+b^2}]$

$=\sqrt{\frac{(a^2+b^2{)}^2-(a^2-b^2{)}^2}{(a^2+b^2{)}^2}}$

$=\sqrt{\frac{(a^2+b^2+a^2-b^2)(a^2+b^2-a^2+b^2)}{a^2+b^2}}$

[Using $x^2-y^2=(x-y)(x+y)]$

$=\sqrt{\frac{2a^2\times 2b^2}{a^2+b^2}}$

$=\frac{2ab}{a^2+b^2}....\left(ii\right)$

Now $tan\theta =\frac{sin\theta }{cos\theta }$

$=\frac{\frac{a^2-b^2}{a^2+b^2}}{\frac{2ab}{a^2+b^2}}$

$=\frac{a^2-b^2}{2ab}$

$sec\theta =\frac{1}{cos\theta }=\frac{a^2+b^2}{2ab}\left[from\right(ii\left)\right]$

and $cosec\theta =\frac{1}{sin\theta }=\frac{a^2+b^2}{a^2-b^2}$ [from (i)]