Question
यदि $\tan \alpha = {(1 + {2^{ - x}})^{ - 1}},$ $\tan \beta = {(1 + {2^{x + 1}})^{ - 1}}$, तब $\alpha + \beta =$
==> $\tan (\alpha + \beta ) = \frac{{\frac{1}{{1 + \frac{1}{{{2^x}}}}} + \frac{1}{{1 + {2^{x + 1}}}}}}{{1 - \frac{1}{{1 + 1/{2^x}}}\frac{1}{{1 + {2^{x + 1}}}}}}$
==> $\tan (\alpha + \beta ) = \frac{{{2^x} + {{2.2}^{x + x}} + {2^x} + 1}}{{1 + {2^x} + {{2.2}^x} + {{2.2}^{x + x}} - {2^x}}}$
==> $\tan (\alpha + \beta ) = 1$
$ \Rightarrow \alpha + \beta = \frac{\pi }{4}$.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.