$$ \text { If } \tan \theta=\frac{a}{b} \text {, then } b \cos 2 \theta+a \sin 2 \theta \text { is equal to } $$
If for real values of $x, \cos \theta=x+\frac{1}{x}$, then
60 The value of $\frac{\sin 50 \Upsilon}{\sin 130 \Upsilon}$ is .............. .
$$\begin{aligned} \text{Here,}\quad \frac{\sin 50 \Upsilon}{\sin 130 \Upsilon} & =\frac{\sin (180 \Upsilon-130 \Upsilon)}{\sin 130 \Upsilon} \\ & =\frac{\sin 130 \Upsilon}{\sin 130 \Upsilon}=1 \end{aligned} $$
If $k=\sin \left(\frac{\pi}{18}\right) \sin \left(\frac{5 \pi}{18}\right) \sin \left(\frac{7 \pi}{18}\right)$, then the numerical value of $k$ is .................. .
Here,
$$\begin{aligned} &\begin{aligned} & k=\sin \left(\frac{\pi}{18}\right) \sin \left(\frac{5 \pi}{18}\right) \sin \left(\frac{7 \pi}{18}\right) \\ & =\sin 10 \Upsilon \sin50\Upsilon \sin70\Upsilon \\ & =\sin 10 \Upsilon \cos 40 r \cdot \cos 20 \Upsilon \\ & =\frac{1}{2} \sin 10 \Upsilon[2 \cos 40 \Upsilon \cdot \cos 20 \Upsilon] \\ & =\frac{1}{2} \sin 10 \Upsilon[\cos 60 \Upsilon+\cos 20 \Upsilon] \quad[\because 2 \cos x \cdot \cos y=\cos (x+y)+\cos (x-y)] \\ & =\frac{1}{2} \sin 10 r \cdot \frac{1}{2}+\frac{1}{2} \sin 10 \Upsilon \cos 20 \Upsilon \\ & =\frac{1}{4} \sin 10 \Upsilon+\frac{1}{4}[\sin 30 \Upsilon-\sin 10 \Upsilon] \\ & =\frac{1}{8} \end{aligned}\\ \end{aligned}$$
$$\text { If } \tan A=\frac{1-\cos \theta}{\sin B} \text {, then } \tan 2 A= ........... .$$
Given that,
$$\begin{aligned} \tan A & =\frac{1-\cos B}{\sin B} \\ & =\frac{1-1+2 \sin ^2 \frac{B}{2}}{2 \sin \frac{B}{2} \cdot \cos \frac{B}{2}}=\tan \frac{B}{2} \end{aligned}$$$$ \begin{array}{ll} \text { Now, } & \tan 2 A=\frac{2 \tan A}{1-\tan ^2 A} \\ \Rightarrow & \tan 2 A=\frac{2 \cdot \tan \frac{B}{2}}{1-\tan ^2 \frac{B}{2}} \\ \Rightarrow & \tan 2 A=\tan B \end{array}$$