Question

If a^2sec^2alpha - b^2tan^2alpha =c^2, then prove that sinalpha = plus(+)minus(-)√c^2-a^2/c^2-b^2​

Answer 1

★Given:-

$\bullet\sf a^2sec^2\alpha-b^2tan^2\alpha=c^2$

★To Prove :-

$\bullet\sf sin\alpha= \pm \sqrt{\dfrac{c^2-a^2}{c^2-b^2}}$

★Solution:-

$\longrightarrow\sf (a\ sec\alpha)^2- (b\ tan\alpha)^2\\ \\ \\ \longrightarrow\sf \bigg(\dfrac{a}{cos\alpha}\bigg)^2-\bigg(\dfrac{bsin\alpha}{cos\alpha}\bigg)^2=c^2 \ \ \ \ \ \ \bigg[\therefore\ sec\theta= \dfrac{1}{Cos\theta}\ \ ; \ \ tan\theta= \dfrac{sin\theta}{cos\theta}\bigg]\\ \\ \\ \sf \ \ \ \ \ Now\ , \ \ \\ \\ \longrightarrow\sf \dfrac{a^2}{cos^2\alpha}-\dfrac{b^2\ sin^2\alpha}{cos^2\alpha}=c^2 \\ \\ \\ \longrightarrow\sf \dfrac{a^2-b^2 \ sin^2\alpha}{cos^2\alpha}=c^2\\ \\ \\ \longrightarrow\sf a^2-b^2\ sin^2\alpha=c^2\ cos^2\alpha\\ \\ \\ \longrightarrow\sf a^2-b^2\sin^2\alpha=c^2(1-sin^2\alpha)\ \ \ \ \ \Big[cos^2\theta= (1-sin^2\theta)\Big]\\ \\ \\ \longrightarrow\sf a^2-b^2\ sin^2\alpha= c^2-c^2sin^2\alpha\\ \\ \\ \longrightarrow\sf \Big(-b^2\ sin^2\alpha+c^2\ sin^2\alpha\Big)= c^2-a^2\\ \\ \\ \longrightarrow\sf sin^2\alpha(-b^2+c^2)= c^2-a^2\\ \\ \\ \longrightarrow\sf sin^2\alpha= \dfrac{c^2-a^2}{c^2-b^2}\\ \\ \\ \longrightarrow\sf sin\alpha= \pm \sqrt{\dfrac{c^2-a^2}{c^2-b^2}}$

$\underline{\bigstar{\sf \ \ Hence \ \ Proved \ !! }}$