JEE Main 2020 — Trigonometry

If sin A + sin B = 1 and cos A + cos B = 0 then the value of 12cos2A + 4cos2B is:

If $\frac{\sqrt{2}sin\alpha }{\sqrt{1+cos2\alpha }}=\frac{1}{7}$ and $\sqrt{\frac{1-cos2\beta }{2}}=\frac{1}{\sqrt{10}},\alpha , \beta \in (0,\frac{\pi }{2})$, then $tan(\alpha +2\beta )$, is equal to

If the equation ${\mathrm{cos}}^{4}\theta +{\mathrm{sin}}^{4}\theta +\lambda =0$ has real solutions for $\theta$ then $\lambda$ lies in interval

If $L={\mathrm{sin}}^{2}(\frac{\pi }{16})-{\mathrm{sin}}^{2}(\frac{\pi }{8})$ and $M={\mathrm{cos}}^{2}(\frac{\pi }{16})-{\mathrm{sin}}^{2}(\frac{\pi }{8})$

If $S$ is the sum of the first 10 terms of the series, ${\mathrm{tan}}^{-1}(\frac{1}{3})+{\mathrm{tan}}^{-1}(\frac{1}{7})+{\mathrm{tan}}^{-1}(\frac{1}{13})+{\mathrm{tan}}^{-1}(\frac{1}{21})+\ldots \ldots$ then $\mathrm{tan}(S)$ is equal to :

The value of ${\mathrm{cos}}^{3}(\frac{\pi }{8}).\mathrm{cos}(\frac{3\pi }{8})+{\mathrm{sin}}^{3}(\frac{\pi }{8}).\mathrm{sin}(\frac{3\pi }{8})$ is:

If $y=\sum _{k=1}^{6}k{\mathrm{cos}}^{-1}{\frac{3}{5}\mathrm{cos}kx-\frac{4}{5}\mathrm{sin}kx}$ then $\frac{dy}{dx}$ at $x=0$is

$2\pi -({\mathrm{sin}}^{-1}\frac{4}{5}+{\mathrm{sin}}^{-1}\frac{5}{13}+{\mathrm{sin}}^{-1}\frac{16}{65})$ is equal to :