$s e c 2 x - \tan 2 x$ in terms of $\tan$ is _________

Open in App

Solution

Finding the value of $s e c 2 x - \tan 2 x$ in terms of $\tan$ is;
$\sec 2 x - \tan 2 x = \frac{1}{\cos 2 x} - \frac{\sin 2 x}{\cos 2 x} = \frac{(1 - \sin 2 x)}{\cos 2 x} = \frac{(\sin^{2} x + \cos^{2} x - 2 sinAcosA)}{(\cos 2 x - \sin 2 x)} [\sin^{2} A + \cos^{2} A = 1, \cos 2 A = \cos^{2} A - \sin^{2} A, \sin 2 A = 2 sinAcosA] = \frac{{(cosx - sinx)}^{2}}{(cosx + sinx) (cosx - sinx)} [0 < x < π / 4 then sinx < cosx] = \frac{(cosx - sinx)}{(cosx + sinx)} = \frac{cosx [1 - (sinx / cosx)]}{cosx [1 + (sinx / cosx)]} = \frac{(1 - tanx)}{(1 + tanx)} = \frac{(tanπ / 4 - tanx)}{(1 + tanx . 1)} = \frac{(tanπ / 4 - tanx)}{(1 + tanxtanπ / 4)} = \tan (π / 4 - x)$
Hence, the answer is $\tan (π / 4 - x)$ .

Suggest Corrections

Similar questions

\int \frac{{sec}^{2} x}{c o {sec}^{2} x} d x

\int \frac{sec 2 x - 1}{sec 2 x + 1} d x

\frac{e^{2 x} - 1}{e^{2 x} + 1}

\int {tan}^{3} 2 x sec 2 x d x = \frac{1}{k} ({sec}^{3} 2 x - 3 sec 2 x) .

k

Join BYJU'S Learning Program