Question

Pls solve this also (°-°)​

Answer 1

Solution :-

Here , in the given question substituting , cos2x = cos²x - sin²x

We have ,

• cos²x - sin²x = $\dfrac{\textbf{\textsf{1-tan ^2 x}}}{\small\textbf{\textsf{1+tan ^2 x}}}$

Taking RHS and substituting

• tanx = $\dfrac{\textbf{\textsf{sinx}}}{\textbf{\textsf{cosx}}}$

$\displaystyle\sf \longrightarrow \dfrac{1-\frac{sin^2x}{cos^2x}}{1+\frac{sin^2x}{cos^2x}}\\\\\displaystyle\sf\longrightarrow \dfrac{\frac{cos^2x-sin^2x}{\cancel{cos^2x}}}{\frac{cos^2x+sin^2x}{\cancel{cos^2x}}}\\\\\displaystyle\sf\longrightarrow \dfrac{cos^2x-sin^2x}{cos^2x + sin^2x}$

Substituting ,

• cos²x + sin²x = 1

$\sf\longrightarrow \dfrac{cos^2x-sin^2x}{1}$

$\longrightarrow \textbf{\textsf{cos ^2 x-sin ^2 x}}$

Now , comparing with RHS

$\longrightarrow$ cos²x - sin²x = cos²x - sin²x

• LHS = RHS

Hence , proved

Answer 2

♣ Qᴜᴇꜱᴛɪᴏɴ :

• Prove : $\sf{cos2x=\:\dfrac{1-tan^2x}{1+tan^2x}}$

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♣ ᴀɴꜱᴡᴇʀ :

$\cos \left(2x\right)=\dfrac{1-\tan ^2\left(x\right)}{1+\tan ^2\left(x\right)}$

$\mathrm{Manipulating\:right\:side}$

$\dfrac{1-\tan ^2\left(x\right)}{1+\tan ^2\left(x\right)}$

$\text { Express with } \sin , \text { cos }$

$=\dfrac{-\left(\dfrac{\sin \left(x\right)}{\cos \left(x\right)}\right)^2+1}{\left(\dfrac{\sin \left(x\right)}{\cos \left(x\right)}\right)^2+1}$

$\text { Simplify } \dfrac{-\left(\dfrac{\sin (x)}{\cos (x)}\right)^{2}+1}{\left(\dfrac{\sin (x)}{\cos (x)}\right)^{2}+1}: \dfrac{-\sin ^{2}(x)+\cos ^{2}(x)}{\sin ^{2}(x)+\cos ^{2}(x)}$

$=\dfrac{\cos ^2\left(x\right)-\sin ^2\left(x\right)}{\cos ^2\left(x\right)+\sin ^2\left(x\right)}$

$\mathrm{Use\:the\:following\:identity}:\quad \cos ^2\left(x\right)+\sin ^2\left(x\right)=1$

$=\dfrac{\cos ^2\left(x\right)-\sin ^2\left(x\right)}{1}$

$=\cos ^2\left(x\right)-\sin ^2\left(x\right)$

$\mathrm{Use\:the\:following\:identity}:\quad \cos ^2\left(x\right)-\sin ^2\left(x\right)=\cos \left(2x\right)$

$=\cos \left(2x\right)$

L.H.S = R.H.S

Hence Proved !!!