MCQ
Write the function in the simplest form: $\tan ^{-1}\left(\frac{\cos x-\sin x}{\cos x+\sin x}\right)$
- A$-\frac{\pi}{4}+x$
- B$-\frac{\pi}{4}-x$
- ✓$\frac{\pi}{4}-x$
- D$\frac{\pi}{4}+x$
$=\tan ^{-1}\left(\frac{1-\left(\frac{\sin x}{\cos x}\right)}{1+\left(\frac{\sin x}{\cos x}\right)}\right)$
$=\tan ^{-1}\left(\frac{1-\tan x}{1+\tan x}\right)$
$=\tan ^{-1}(1)-\tan ^{-1}(\tan x)$ $\left[\because \frac{-y}{x y}=\tan ^{-1} x-\tan ^{-1} y\right]$
$=\frac{\pi}{4}-x$
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$(A)$ $\frac{1}{\mathrm{PS}}+\frac{1}{\mathrm{ST}}<\frac{2}{\sqrt{\mathrm{QS} \times \mathrm{SR}}}$
$(B)$ $\frac{1}{\mathrm{PS}}+\frac{1}{\mathrm{ST}}>\frac{2}{\sqrt{\mathrm{QS} \times \mathrm{SR}}}$
$(C)$ $\frac{1}{\mathrm{PS}}+\frac{1}{\mathrm{ST}}<\frac{4}{\mathrm{QR}}$
$(D)$ $\frac{1}{\mathrm{PS}}+\frac{1}{\mathrm{ST}}>\frac{4}{\mathrm{QR}}$
$(A)$ Projection of $\overline{ OC }$ on $\overline{ OA }$ is $-\frac{3}{2}$
$(B)$ Area of the triangle $OAB$ is $\frac{9}{2}$
$(C)$ Area of the triangle $ABC$ is $\frac{9}{2}$
$(D)$ The acute angle between the diagonals of the parallelogram with adjacent sides $\overline{ OA }$ and $\overline{ OC }$ is $\frac{\pi}{3}$