Question

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Answer 1

$\underline{\textbf{QUESTION :-}}$

$\textsf{If cosec \:0 + cot \:0 =p, then prove that cos \:0}$ $= \frac{p^2 -1}{ p^2+1}$

$\underline{\textbf{SOLUTION :-}}$

Given :$cosec0+cot0=p$

To Prove:$cos0=\frac{p^2-1}{p^2+1}$

Proof - $Cosec0+cot0=p$ $\to (i)$

We know-$cosec^20-cot^20=1$

$(cosec \:0+cot\:0)(cosec\:0+cot\:0)=1$

$cosec0-cot0=\frac{1}{p}\to (ii)$ $\textsf{from i}$

Add (i) & (ii)

$cosec\:0+cot\:0+cosec\:0-cot\:0=p+\frac{1}{p}$

$2 \:cosec\:0=\frac{p^2+1}{p}\to(iii)$

Subtract  (i) & (ii)

$cosec\:0+cot\:0-cosec\:0+cot\:0=p-\frac{1}{p}$

$2\:cot\:0=\frac{p^2-1}{p}\to (iv)$

Divide $\frac{iv}{iii}$

$\frac{2\:cot\:0}{2\:cosec\:0}=\frac{p^2-1}{p^2+1}$

$\frac{cos\:o\times sin\:o}{sin\:o}=\frac{p^2-1}{p^2+1}$

$cos0=\frac{p^2-1}{p^2+1}$

$\boxed{cos\:0=\frac{p^2-1}{p^2+1}}$

$\therefore\: \mathcal{HENCE\:PROVED}$

Answer 2

Given :

• $\displaystyle{\sf cosec\theta+cot\theta=p}$

To Prove :

• $\displaystyle{\sf cos\theta = \dfrac{p^2-1}{p^2+1}}$

Solution :

${\longrightarrow \displaystyle{\sf cosec\theta+cot\theta=p}}$

Substitute the value of p in the Equation.

${\longrightarrow\displaystyle{\sf \dfrac{p^2-1}{p^2+1}}}\\\\{\longrightarrow \displaystyle{\sf \dfrac{(cosec\theta+cot\theta)^2-1}{(cosec\theta+cot\theta)^2+1}}}\\\\{\longrightarrow \displaystyle{\sf \dfrac{cosec^2\theta+cot^2\theta+2cosec\theta\:cot\theta-1}{cosec^2\theta+cot^2\theta+2cosec\theta\:cot\theta+1}}}\\\\{\longrightarrow \displaystyle{\sf \dfrac{(cosec\theta^2-1)+cot\theta^2+2cosec\theta\:cot\theta}{(cot^2\theta+1)+cosec^2\theta+2cosec\theta\:cot\theta}}}\\\\{\longrightarrow \displaystyle{\sf \dfrac{cot^2\theta+cot^2\theta+2cosec\theta\:cot\theta}{cosec^2\theta+cosec^2\theta+2cosec\theta\:cot\theta}}}\\\\{\longrightarrow \displaystyle{\sf \dfrac{2cot^2\theta+2cosec\theta\:cot\theta}{2cosec^2\theta+2cosec\theta\:cot\theta}}}\\\\{\longrightarrow \displaystyle{\sf \dfrac{2cot\theta{\cancel{(cot\theta+cosec\theta)}}}{2cosec\theta{\cancel{(cosec\theta+\:cot\theta)}}}}}\\\\{\longrightarrow \displaystyle{\sf \dfrac{{\cancel{2}}cot\theta}{{\cancel{2}}cosec\theta}}}\\\\{\longrightarrow \displaystyle{\sf \dfrac{cot\theta}{cosec\theta}}}\\\\{\longrightarrow \displaystyle{\sf \dfrac{cos\theta}{{\cancel{sin\theta}}}\times \dfrac{{\cancel{sin\theta}}}{1}}}\\\\{\longrightarrow \displaystyle{\sf cos\theta }}\\\\{\sf {Hence\:Proved}}$