(1)      $\frac{7}{25}$

(1)    1/6

(2)     30°

(2)

$\sin \theta +\cos \theta =\sqrt{2}$

Squaring on both sides, we get

${\sin }^2\theta +{\cos }^2\theta +2\sin \theta \cos \theta =2$

$\Rightarrow 1+2\sin \theta \cos \theta =2$  $[\because {\sin }^2\theta +{\cos }^2\theta =1]$

$\Rightarrow 2\sin \theta \cos \theta =1\Rightarrow \sin \theta \cos \theta =\frac{1}{2}$

But $\tan \theta +\cot \theta =\frac{\sin \theta }{\cos \theta }+\frac{\cos \theta }{\sin \theta }=\frac{{\sin }^2\theta +{\cos }^2\theta }{\cos \theta \sin \theta }=2$

(1)

$5\tan \beta =4 \Rightarrow \tan \beta =\frac{4}{5}$

$\therefore$ $\frac{5\sin \beta -2\cos \beta }{5\sin \beta +2\cos \beta }=\frac{5\tan \beta -2}{5\tan \beta +2}$     $\text{[dividing }\cos \beta \text{ by Nr. and Dr.]}$

                                    $=\frac{5\times \frac{4}{5}-2}{5\times \frac{4}{5}+2}=\frac{2}{6}=\frac{1}{3}$

(2)

$x\tan 60°\cos 60°=\sin 60°\cot 60°$

$\Rightarrow x\times \sqrt{3}\times \frac{1}{2}=\frac{\sqrt{3}}{2}\times \frac{1}{\sqrt{3}}\Rightarrow x\sqrt{3}=1$

$\Rightarrow x=\frac{1}{\sqrt{3}}\Rightarrow x=\tan 30°$ $[\because \tan 30°=\frac{1}{\sqrt{3}}]$

(3)

Given, $\sec \theta -\tan \theta =m$ ...(i)

We know that, ${\sec }^2\theta -{\tan }^2\theta =1$

$\Rightarrow (\sec \theta -\tan \theta )(\sec \theta +\tan \theta )=1$

$\Rightarrow m(\sec \theta +\tan \theta )=1$

$\Rightarrow \sec \theta +\tan \theta =\frac{1}{m}$

(1)

$\cos (\alpha +\beta )=0=\cos 90°\Rightarrow (\alpha +\beta )=90°\Rightarrow \frac{\alpha +\beta }{2}=45°$

$\Rightarrow \cos \left(\frac{{\displaystyle \alpha +\beta }}{{\displaystyle 2}}\right)=\cos 45°=\frac{1}{\sqrt{2}}$

(3)     1

(1)       Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A)