Question

question in attachment​

Answer 1

Given :-

$\rm( 1+ tan^{2} A ) + \bigg( 1 + \dfrac{1}{tan^{2} A} \bigg) = \dfrac{1}{ {sin}^{2} A - {sin}^{4} A}$

To prove :-

• LHS = RHS

Proof :-

$\rm \:LHS = \rm( 1+ tan^{2} A ) + \bigg( 1 + \dfrac{1}{tan^{2} A} \bigg) \\ \\ \rm RHS = \dfrac{1}{ {sin}^{2} A - {sin}^{4} A}$

$\sf LHS = \rm( 1+ tan^{2} A ) + \bigg( 1 + \dfrac{1}{tan^{2} A} \bigg) \\ \\ \longrightarrow \rm( 1+ \frac{ {sin}^{2} A}{ {cos}^{2} A} ) + \bigg( 1 + \dfrac{{cos}^{2} A}{{sin}^{2} A} \bigg)\\ \\ \\ \longrightarrow \rm( \dfrac{ {cos}^{2} A + {sin}^{2} A}{ {cos}^{2} A} ) + \bigg( \dfrac{{cos}^{2} A + {sin}^{2} A}{{sin}^{2} A} \bigg)\\ \\ \\ \longrightarrow \rm \bigg( \dfrac{1}{{cos}^{2} A} \bigg)+ \bigg( \dfrac{1}{{sin}^{2} A} \bigg)\\ \\ \\ \longrightarrow \rm \dfrac{{cos}^{2} A +{sin}^{2} A }{{cos}^{2}A \: {sin}^{2} A} = \dfrac{1}{{cos}^{2} A \: {sin}^{2} A}......(1)$

$\rm RHS = \dfrac{1}{ {sin}^{2} A - {sin}^{4} A} \\ \\ \\ \rm \longrightarrow \dfrac{1}{ {sin}^{2} A - {sin}^{4} A}\\ \\ \\ \rm \longrightarrow \dfrac{1}{ {sin}^{2} A (1- {sin}^{2} A)}\\ \\ \\ \rm \longrightarrow \dfrac{1}{ {sin}^{2} A \: {cos}^{2} A}.......(2)$

Therefore , from (1) and (2) we get :- LHS = RHS

hence proved!

• Additional Information :-

• Trigonometric Identities :-

$\boxed{\begin{minipage}{6cm} Important Trigonometric identities :- \\ \\ \: \: 1)\:\sin^2\theta+\cos^2\theta=1 \\ \\ 2)\:\sin^2\theta= 1-\cos^2\theta \\ \\ 3)\:\cos^2\theta=1-\sin^2\theta \\ \\ 4)\:1+\cot^2\theta=\text{cosec}^2 \, \theta \\ \\5)\: \text{cosec}^2 \, \theta-\cot^2\theta =1 \\ \\ 6)\:\text{cosec}^2 \, \theta= 1+\cot^2\theta \\\ \\ 7)\:\sec^2\theta=1+\tan^2\theta \\ \\ 8)\:\sec^2\theta-\tan^2\theta=1 \\ \\ 9)\:\tan^2\theta=\sec^2\theta-1 \end{minipage}}$

Answer 2

${\huge{\underline{\boxed{\red{\mathcal{Answer}}}}}}$

To Prove:

• $\left(1 + \tan^2A \right) + \left(1 + \dfrac{1}{\tan^2A} \right) = \dfrac{1}{\sin^2A - \sin^4A}$

Proof:

LHS = $\left(1 + \tan^2A \right) + \left(1 + \dfrac{1}{\tan^2A} \right)$

RHS = $\dfrac{1}{\sin^2A - \sin^4A}$

$Taking\:LHS, \implies \left(1 + \tan^2A \right) + \left(1 + \dfrac{1}{\tan^2A} \right) \\ \\ \implies \left(1 + \dfrac{\sin^2A}{\cos^2A} \right) + \left(1 + \dfrac{\cos^2A}{\sin^2A} \right) \\ \\ \implies \left(\dfrac{\cos^2A+\sin^2A}{\cos^2A} \right) + \left(\dfrac{\sin^2A+\cos^2A}{\sin^2A} \right) \\ \\ \implies \left(\dfrac{1}{\cos^2A} \right) + \left(\dfrac{1}{sin^2A} \right) \\ \\ \implies \left(\dfrac{\sin^2A+\cos^2A}{\sin^2A*\cos^2A} \right) \\ \\ \implies \left(\dfrac{1}{\sin^2A*\cos^2A} \right) \\ \\ \implies \left(\dfrac{1}{\sin^2A*(1-\sin^2A)} \right) \\ \\ \implies \left(\dfrac{1}{\sin^2A - \sin^4A}\right) = RHS\: \\ \\ \\ HENCE\:PROVED$

Formula used:

• $\sin^2A + \cos^2A = 1$
• $\cos^2A = 1 - \sin^2A$
• $\tan A = \dfrac{\sin A}{\cos A}$

Learn More:

$\boxed{\begin{minipage}{6cm} Important Trigonometric identities :- \\ \\ \: \: 1)\:\sin^2\theta+\cos^2\theta=1 \\ \\ 2)\:\sin^2\theta= 1-\cos^2\theta \\ \\ 3)\:\cos^2\theta=1-\sin^2\theta \\ \\ 4)\:1+\cot^2\theta=\text{cosec}^2 \, \theta \\ \\5)\: \text{cosec}^2 \, \theta-\cot^2\theta =1 \\ \\ 6)\:\text{cosec}^2 \, \theta= 1+\cot^2\theta \\\ \\ 7)\:\sec^2\theta=1+\tan^2\theta \\ \\ 8)\:\sec^2\theta-\tan^2\theta=1 \\ \\ 9)\:\tan^2\theta=\sec^2\theta-1 \end{minipage}}$

HOPE IT HELPS!!!