1.

Prove that (2+2secθ)(1−secθ)/(2+2cosecθ)(1−cosecθ)=tan4 θPLs ANSWER FAST I WILL MARK AS BRAINLIEST​

Answer»

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Given :-

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

To Prove :-

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

Formula APPLIED :-

\begin{gathered}\bullet\ \; \sf sec^2\theta-tan^2\theta=1\\\\\to\ \sf -tan^2\theta=1-sec^2\theta\end{gathered}

\begin{gathered}\bullet\ \; \ \sf csc^2\theta-cot^2\theta=1\\\\\to\ \; \sf -cot^2\theta=1-csc^2\theta\end{gathered}

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

\bold{∙(a+b)\:(a−b)\:=\:a² - b²}

PROOF :

TAKE LHS

\begin{gathered}\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)}\end{gathered}

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

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

\begin{gathered}\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}}\end{gathered}

\bold{→\:tan² θ.tan²θ}

\bold{→tan⁴ θ}

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RHS

More INFO :-

\begin{gathered}\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}\end{gathered}

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