Question

please solve this question anyone ​

Answer 1

Solution

To Prove

$\sf \: \dfrac{1}{ {sin}^{2} \theta } - \dfrac{1}{ {sin}^{2} \phi} = \dfrac{ {cos}^{2} \theta - {cos}^{2} \phi}{ {sin}^{2} \theta.sin {}^{2} \phi }$

LHS

$\sf \: \dfrac{1}{ {sin}^{2} \theta } - \dfrac{1}{ {sin}^{2} \phi} \\ \\ \dashrightarrow \: \sf \: \dfrac{sin {}^{2} \phi \: - {sin}^{2} \theta}{sin {}^{2} \theta \: {sin}^{2} \phi }$

Since,

sin²∅ + cos²∅ = 1

➠ sin²∅ = 1 - cos²∅

Thus,

$\dashrightarrow \: \sf \: \dfrac{(1 - {cos}^{2} \phi) - ( 1 - {cos}^{2} \theta)}{sin {}^{2} \theta \: . {sin}^{2} \phi } \\ \\ \dashrightarrow \: \sf \: \dfrac{ {cos}^{2} \theta \: - {cos}^{2} \phi \: + 1 - 1 }{ {sin}^{2} \theta . {sin}^{2} \phi} \\ \\ \dashrightarrow \: \sf \: \frac{ {cos}^{2} \theta - cos {}^{2} \phi }{ {sin}^{2} \theta. {sin}^{2} \phi}$

Henceforth, ProveD

Answer 2

❏ Question:-

@ Prove that

$\implies\sf{\ \ {\dfrac{1}{\sin^2\theta}-\dfrac{1}{\sin^2\phi}=\dfrac{\cos^2\theta-\cos^2\phi}{\sin^2\theta\sin^2\phi}}}$

❏ Solution:-

$\blacksquare\:\:\:\sf{\ \ {L.H.S.=\dfrac{1}{\sin^2\theta}-\dfrac{1}{\sin^2\phi}}}$

$\implies\sf{\ \ {L.H.S.=\dfrac{\sin^2\phi-\sin^2\theta}{\sin^2\theta\sin^2\phi}}}$

$\implies\sf{\ \ {L.H.S.=\dfrac{(1-\cos^2\phi)-(1-\cos^2\theta)}{\sin^2\theta\sin^2\phi}}}$

$\implies\sf{\ \ {L.H.S.=\dfrac{\cancel1-\cos^2\phi-\cancel1+\cos^2\theta}{\sin^2\theta\sin^2\phi}}}$

$\implies\sf{\ \ {L.H.S.=\dfrac{\cos^2\theta-\cos^2\phi}{\sin^2\theta\sin^2\phi}}}$

$\implies\boxed{\sf{\ \ {L.H.S.=R.H.S.}}}$ (proved)

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