$ = \tan \left[ {\frac{1}{2}{{\sin }^{ - 1}}\left( {\frac{{2\tan \theta }}{{1 + {{\tan }^2}\theta }}} \right) + \frac{1}{2}{{\cos }^{ - 1}}\left( {\frac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }}} \right)} \right]$
(Let $a = \tan \theta $)
$ = \tan \left[ {\frac{1}{2}{{\sin }^{ - 1}}(\sin 2\theta ) + \frac{1}{2}{{\cos }^{ - 1}}(\cos 2\theta )} \right]$
$ = \tan (2\theta ) = \tan 2\theta = \frac{{2\tan \theta }}{{1 - {{\tan }^2}\theta }} = \frac{{2a}}{{1 - {a^2}}}$
Trick : Put $a = 0$, then tan $(0+0)=0;$ which is given by $(a)$ and $(c)$.
Again put $a = 1$, then $\tan \left( {\frac{\pi }{4} + \frac{\pi }{4}} \right) = \infty $, which is given by $(c).$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$I.$ Adifferentiable function $' f '$ with maximum at $x = c$ ==> $ f "(c) < 0$.
$II.$ Antiderivative of a periodic function is also a periodic function.
$III.$ If $f$ has a period $T$ then for any $a \in R$. $\int\limits_0^T {f(x)\,dx} = \int\limits_0^T {f(x + a)\,dx} $
$IV.$ If $f (x)$ has a maxima at $x = c$ , then $'f '$ is increasing in $(c - h, c)$ and decreasing in $(c, c + h)$ as $h \rightarrow 0$ for $h > 0.$ Now indicate the correct alternative.
$-x+2 y+5 z=b_1$
$2 x-4 y+3 z=b_2$
$x-2 y+2 z=b_3$
has at least one solution. Then, which of the following system(s) (in real variables) has (have) at least one solution for each$\left[\begin{array}{l}b_1 \\ b_2 \\ b_3\end{array}\right]$ $\in$ $S$ ?
$(A)$ $x+2 y+3 z=b_1, 4 y+5 z=b_2$ and $x+2 y+6 z=b_3$
$(B)$ $x+y+3 z=b_1, 5 x+2 y+6 z=b_2$ and $-2 x-y-3 z=b_3$
$(C)$ $-x+2 y-5 z=b_1, 2 x-4 y+10 z=b_2$ and $x-2 y+5 z=b_3$
$(D)$ $x+2 y+5 z=b_1, 2 x+3 z=b_2$ and $x+4 y-5 z=b_3$