Question

cosec 2 theta +sec 2 theta/cosec2theta-sec2theta=1+tan2 theta/1-tan2 theta

Answer 1

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \boxed{\boxed { \huge \mathcal\red{ solution}}}}$

$\underline{\red{\textbf{TO PROVE}}}$

$\implies \boxed{\bf\frac{cosec2\theta+sec2\theta}{cosec2\theta-sec2\theta}=\bf\frac{1+tan2\theta}{1-tan2\theta}}$

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$\star\red{\textbf{Proving}}$

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$\textbf{L.H.S.}=\bf\frac{cosec2\theta+sec2\theta}{cosec2\theta-sec2\theta}$

$=\bf\frac{\frac{1}{sin2\theta}+\frac{1}{cos2\theta}}{\frac{1}{sin2\theta}-\frac{1}{cos2\theta}}$

$=\Large{\bf\frac{\frac{cos2\theta+sin2\theta}{sin2\theta cos2\theta}}{\frac{cos2\theta-sin2\theta}{sin2\theta cos2\theta}}}$

$=\bf\frac{cos2\theta+sin2\theta}{cos2\theta-sin2\theta}\times \frac{\cancel{sin2\theta cos2\theta}}{\cancel{sin2\theta cos2\theta}}$

$=\Large{\bf\frac{\frac{cos2\theta+sin2\theta}{cos2\theta}}{\frac{cos2\theta-sin2\theta}{cos2\theta}}}\\ [\blue{\bf\textit{dividing by}\:cos2\theta\textit{on both sides}}]$

$=\Large{\bf\frac{\frac{\cancel{cos2\theta}}{\cancel{cos2\theta}}+\frac{sin2\theta}{cos2\theta}}{\frac{\cancel{cos2\theta}}{\cancel{cos2\theta}}-\frac{sin2\theta}{cos2\theta}}}$

$=\Large{\bf\frac{1+tan2\theta}{1-tan2\theta}=\textbf{R.H.S.}}\:\:(proved)$

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$\underline{ \huge\mathfrak{hope \: this \: helps \: you}}$

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Answer 2

To prove that: $\dfrac{\csc^ 2 \theta +\sec^ 2 \theta}{\csc^ 2 \theta -\sec^ 2 \theta} =\dfrac{1+\tan^2 \theta}{1-\tan^2 \theta} .$

L.H.S. = $\dfrac{\csc^ 2 \theta +\sec^ 2 \theta}{\csc^ 2 \theta -\sec^ 2 \theta}$

Using the trigonometric identity:

$\csc^ 2 \theta$ = $\dfrac{1}{\sin^ 2 \theta }$  and $\sec^ 2 \theta = \dfrac{1}{\cos^ 2 \theta}$

= $\dfrac{\dfrac{1}{\sin^ 2 \theta } +\dfrac{1}{cos^ 2 \theta} }{\dfrac{1}{\sin^ 2 \theta } -\dfrac{1}{cos^ 2 \theta}}$

Taking LCM of denominator part, we get

$=\dfrac{\dfrac{\cos^ 2 \theta+\sin^ 2 \theta}{\sin^ 2 \theta\cos^ 2 \theta } }{\dfrac{\cos^ 2 \theta-\sin^ 2 \theta}{\sin^ 2 \theta\cos^ 2 \theta }}$

$=\dfrac{\cos^ 2 \theta+\sin^ 2 \theta }{\cos^ 2 \theta-\sin^ 2 \theta}$

Using the trigonometric identity:|

$\cos^ 2 \theta+\sin^ 2 \theta$ = 1

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

R.H.S. = $\dfrac{1+\tan^2 \theta}{1-\tan^2 \theta}$

= $\dfrac{1+\dfrac{\sin^2 \theta}{\cos^2 \theta} }{1-\dfrac{\sin^2 \theta}{\cos^2 \theta}}$

$=\dfrac{\dfrac{\cos^ 2 \theta+\sin^ 2 \theta}{\cos^ 2 \theta } }{\dfrac{\cos^ 2 \theta-\sin^ 2 \theta}{\cos^ 2 \theta }}$

$=\dfrac{\cos^ 2 \theta+\sin^ 2 \theta }{\cos^ 2 \theta-\sin^ 2 \theta}$

Using the trigonometric identity:|

$\cos^ 2 \theta+\sin^ 2 \theta$ = 1

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

∴ L.H.S. = R.H.S. $=\dfrac{1 }{\cos^ 2 \theta-\sin^ 2 \theta}$, proved.

Thus, $\dfrac{\csc^ 2 \theta +\sec^ 2 \theta}{\csc^ 2 \theta -\sec^ 2 \theta} =\dfrac{1+\tan^2 \theta}{1-\tan^2 \theta}$, proved.