Math, asked by saimohitkarthikeyay, 8 months ago

Prove that (2+2secθ)(1−secθ)/(2+2cosecθ)(1−cosecθ)=tan4 θ

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Answers

Answered by BrainlyIAS
38

Given :

\bullet\ \; \sf \dfrac{(2+2sec\theta)(1-sec\theta)}{(2+2csc\theta)(1-csc\theta)}=tan^4\theta

To Prove :

\bullet\ \; \sf \dfrac{(2+2sec\theta)(1-sec\theta)}{(2+2csc\theta)(1-csc\theta)}=tan^4\theta

Formula Applied :

\bullet\ \; \sf sec^2\theta-tan^2\theta=1\\\\\to\  \sf -tan^2\theta=1-sec^2\theta

\bullet\ \;  \ \sf csc^2\theta-cot^2\theta=1\\\\\to\ \; \sf -cot^2\theta=1-csc^2\theta

\bullet\ \; \sf cot \theta=\dfrac{1}{tan\theta}\\\\\to \sf cot^2\theta=\dfrac{1}{tan^2\theta}

\sf \bullet\ \; (a+b)(a-b)=a^2-b^2

Proof :

Take LHS

\to \sf \dfrac{(2+2sec\theta)(1-sec\theta)}{(2+2csc\theta)(1-csc\theta)}\\\\\to \sf \dfrac{2(1+sec\theta)(1-sec\theta)}{2(1+csc\theta)(1-csc\theta)}

\to \sf \dfrac{2(1^2-sec^2\theta)}{2(1^2-csc^2\theta)}\\\\

\to \sf \dfrac{1-sec^2\theta}{1-csc^2\theta}

\to \sf \dfrac{-tan^2\theta}{-cot^2\theta}\\\\\to\ \sf \dfrac{tan^2\theta}{cot^2\theta}\\\\\to\ \sf \dfrac{tan^2\theta}{\frac{1}{tan^2\theta}}

\to \sf tan^2\theta.tan^2\theta

\to \sf tan^4\theta

   RHS

More Info :

\begin{gathered}\bullet\:\bf Trigonometric\:Values :\\\\\boxed{\begin{tabular}{c|c|c|c|c|c}Radians/Angle & 0 & 30 & 45 & 60 & 90\\\cline{1-6}Sin \theta & 0 & $\dfrac{1}{2} &$\dfrac{1}{\sqrt{2}} & $\dfrac{\sqrt{3}}{2} & 1\\\cline{1-6}Cos \theta & 1 & $\dfrac{\sqrt{3}}{2}&$\dfrac{1}{\sqrt{2}}&$\dfrac{1}{2}&0\\\cline{1-6}Tan \theta&0&$\dfrac{1}{\sqrt{3}}&1&\sqrt{3}&Not D{e}fined\end{tabular}}\end{gathered}

Answered by ZAYNN
40

Answer:

\bf{To\:Prove :}\:\sf\dfrac{(2+2sec\theta)(1-sec\theta)}{(2+2csc\theta)(1-csc\theta)}=tan^4\theta

\bf{Proof :}

:\implies\sf \dfrac{(2+2sec\theta)(1-sec\theta)}{(2+2csc\theta)(1-csc\theta)}\\\\\\:\implies\sf \dfrac{2(1+sec\theta)(1-sec\theta)}{2(1+csc\theta)(1-csc\theta)}\\\\\\:\implies\sf \dfrac{2(1^2-sec^2\theta)}{2(1^2-csc^2\theta)}\\\\\\:\implies\sf \dfrac{-tan^2\theta}{-cot^2\theta}\\\\\\:\implies\sf \dfrac{tan^2\theta}{cot^2\theta}\\\\\\:\implies\sf \dfrac{tan^2\theta}{\frac{1}{tan^2\theta}}\\\\\\:\implies\sf tan^2\theta \times tan^2\theta\\\\\\:\implies\underline{\boxed{\sf tan^4\theta}}

\rule{180}{1.5}

\bigstar\:\sf Trigonometric\:Values :\\\begin{tabular}{|c|c|c|c|c|c|}\cline{1-6}Radians/Angle & 0 & 30 & 45 & 60 & 90\\\cline{1-6}Sin \theta & 0 & $\dfrac{1}{2} &$\dfrac{1}{\sqrt{2}} & $\dfrac{\sqrt{3}}{2} & 1\\\cline{1-6}Cos \theta & 1 & $\dfrac{\sqrt{3}}{2}&$\dfrac{1}{\sqrt{2}}&$\dfrac{1}{2}&0\\\cline{1-6}Tan \theta&0&$\dfrac{1}{\sqrt{3}}&1&\sqrt{3}& Not D$\hat{e}$fined \\\cline{1-6}\end{tabular}

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