Question

Solve above question plz​

Answer 1

$\rule{200}4$

$\dagger \;\underline{\underline{\purple{\small\rm We\;Need\;To \;Prove :-}}}$

• $\small\sf \dfrac{tan^2\theta}{tan^2\theta-1}+\dfrac{cosec^2\theta}{sec^2\theta-cosec^2\theta}=\dfrac{1}{sin^2\theta-cos^2\theta}$

$\dagger \;\underline{\underline{\purple{\small\rm Taking\;LHS :-}}}$

Substituting ,

☞ $\small\sf tan^2\theta = \dfrac{sin^2\theta}{cos^2\theta}$

☞ $\small\sf cosec^2\theta = \dfrac{1}{sin^2\theta}$

☞ $\small\sf sec^2\theta =\dfrac{1}{cos^2\theta}$

$\dashrightarrow\sf \small\dfrac{\frac{sin^2\theta}{\cancel{cos^2\theta}}}{\frac{sin^2\theta-cos^2\theta}{\cancel{cos^2\theta}}}+\dfrac{\frac{1}{\cancel{sin^2\theta}}}{\frac{sin^2\theta-cos^2\theta}{\cancel{sin^2\theta} cos^2\theta}}$

$\dashrightarrow\sf \small \dfrac{sin^2\theta}{sin^2\theta-cos^2\theta}+\dfrac{cos^2\theta}{sin^2\theta-cos^2\theta}$

$\dashrightarrow\sf \small \dfrac{sin^2\theta+cos^2\theta}{sin^2\theta-cos^2\theta}$

$\dashrightarrow \boxed{\red{\sf \small \dfrac{1}{sin^2\theta-cos^2\theta}}}$

$\textbf{\textsf{\pink{Hence proved}}}\green{\boldsymbol\checkmark}$

It can also be simplified

⇒ $\small\sf \dfrac{1}{sin^2\theta-cos^2\theta}$

Taking minus (-) common,

⇒ $\small \sf\dfrac{1}{-\big(cos^2\theta-sin^2\theta\big)}$

⇒ $\small\sf \dfrac{1}{-\big(cos2\theta\big)}$

⇒ $\small\sf -sec2\theta$

$\rule{200}4$

Answer 2

$\tt \frac{{tan}^{2}\theta}{{tan}^{2}\theta - 1 } + \frac{{cosec}^{2}\theta}{{sec}^{2}\theta - {cosec}^{2}\theta} = \frac{1}{{sin}^{2}\theta - {cos}^{2}\theta}\\ \\ \tt L.H.S \\ \\ \tt \frac{{tan}^{2}\theta}{{tan}^{2}\theta - 1 } + \frac{{cosec}^{2}\theta}{{sec}^{2}\theta - {cosec}^{2}\theta}\\ \\ \tt converting\:the\:ratios\:in\:sine\:or\:cosine\:we\:get \\ \\ \tt \frac{\frac{{sin}^{2}\theta}{{cos}^{2}\theta}}{\frac{{sin}^{2}\theta}{{cos}^{2}\theta} - 1 } + \frac{\frac{1}{{sin}^{2}\theta}}{\frac{1}{{cos}^{2}\theta} - \frac{1}{{sin}^{2}\theta}} \\ \\ \tt \frac{\frac{{sin}^{2}\theta}{{cos}^{2}\theta}}{\frac{{sin}^{2}\theta-{cos}^{2}\theta}{{cos}^{2}\theta}} + \frac{\frac{1}{{sin}^{2}\theta}}{\frac{{sin}^{2}\theta -{cos}^{2}\theta}{{cos}^{2}\theta.{sin}^{2}\theta}}\\ \\ \tt \frac{\frac{{sin}^{2}\theta}{\cancel{{cos}^{2}\theta}}}{\frac{{sin}^{2}\theta-{cos}^{2}\theta}{\cancel{{cos}^{2}\theta}}} + \frac{\frac{1}{\cancel{{sin}^{2}\theta}}}{\frac{{sin}^{2}\theta -{cos}^{2}\theta}{{cos}^{2}\theta.\cancel{{sin}^{2}\theta}}}\\ \\ \tt \frac{{sin}^{2}\theta}{{sin}^{2}\theta - {cos}^{2}\theta} + \frac{{cos}^{2}\theta}{{sin}^{2}\theta - {cos}^{2}\theta} \\ \\ \tt \frac{{sin}^{2}\theta +{cos}^{2}\theta}{{sin}^{2}\theta -{cos}^{2}\theta}\\ \\ \boxed{\underline{\red{\tt IDENTITY : {sin}^{2}\theta +{cos}^{2}\theta = 1 }}} \\ \\ \tt \frac{1}{{sin}^{2}\theta - {cos}^{2}\theta} \\ \\ \tt R.H.S \\ \\ \boxed{\underline{\red{\tt Hence\:proved}}} \\ \\ \green{\tt NOTE : If\:you\:are\: getting\:problem\:with\:this\:type\:of\: question\:just }\\ \green{\tt convert\:it\:into\:sine\:or\:cosine\:ratios}$

Hope it helps you

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