We have, \(2\cos { 3\theta } =\sqrt { 3 } \)
\(\Rightarrow \quad \cos { 3\theta } =\frac { \sqrt { 3 } }{ 2 } \)
\(\Rightarrow \cos { 3\theta } =\cos { { 30 }^{ 0 } } \)     \(\left[ \because \cos { { 30 }^{ 0 } } =\frac { \sqrt { 3 } }{ 2 } \right] \)
\(\Rightarrow \quad 3\theta ={ 30 }^{ 0 },\)   as 3\(\theta\) and 30⁰ are are acute angles.
\(\therefore \quad \theta ={ 10 }^{ 0 }\)

since, sin (A - B) = \(\frac{1}{2}\), therefore, A - B = 30°              (1)
Also, since cos (A + B) = \(\frac{1}{2}\), therefore, A + B = 60°   (2)
Solving (1) and (2), we get : A = 45° and B = 15°.

sin 60° cos 30° + sin 30° cos 60°
\(=\frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2}+\frac{1}{2} \times \frac{1}{2}\)
\(\left[\because \sin 60^{\circ}=\cos 30^{\circ}=\frac{\sqrt{3}}{2} \text { and } \sin 30^{\circ}=\cos 60^{\circ}=\frac{1}{2}\right]\)
\(=\frac{3}{4}+\frac{1}{4}=\frac{3+1}{4}=\frac{4}{4}=1\)

2tan²45° + cos²30° − sin²60°
\(=2(1)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}-\left(\frac{\sqrt{3}}{2}\right)^{2}\)
\(=2(1)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}-\left(\frac{\sqrt{3}}{2}\right)^{2}\)
\(=2+\frac{3}{4}-\frac{3}{4}=2\) 

(cos 45°)/(sec 30° + cosec 30°)
\(=\frac{\frac{1}{\sqrt{2}}}{\frac{2}{\sqrt{3}}+2}=\frac{\frac{1}{\sqrt{2}}}{\frac{2+2 \sqrt{3}}{\sqrt{3}}}\)
\(=\frac{\sqrt{3}}{\sqrt{2}(2+2 \sqrt{3})}=\frac{\sqrt{3}}{2 \sqrt{2}+2 \sqrt{6}}\)
\(=\frac{\sqrt{3}(2 \sqrt{6}-2 \sqrt{2})}{((2 \sqrt{6})+2 \sqrt{2})(2 \sqrt{6}-2 \sqrt{2})}\)
\(=\frac{2 \sqrt{3}(\sqrt{6}-\sqrt{2})}{(2 \sqrt{6})^{2}-(2 \sqrt{2})^{2}}=\frac{2 \sqrt{3}(\sqrt{6}-\sqrt{2})}{24-8}=\frac{2 \sqrt{3}(\sqrt{6}-\sqrt{2})}{16}\)
\(=\frac{\sqrt{18}-\sqrt{6}}{8}=\frac{3 \sqrt{2}-\sqrt{6}}{8}\) 

\(\frac{\sin 30^{\circ}+\tan 45^{\circ}-\operatorname{cosec} 60^{\circ}}{\sec 30^{\circ}+\cos 60^{\circ}+\cot 45^{\circ}}\)
\(=\frac{\frac{1}{2}+\frac{1}{1}-\frac{2}{\sqrt{3}}}{\frac{2}{\sqrt{3}}+\frac{1}{2}+\frac{1}{1}}=\frac{\frac{\sqrt{3}+2 \sqrt{3}-4}{2 \sqrt{3}}}{\frac{4+\sqrt{3}+2 \sqrt{3}}{2 \sqrt{3}}}\)
\(\begin{aligned} & {\left[\begin{array}{c} \because \sin 30^{\circ}=\cos 60^{\circ}=\frac{1}{2}, \operatorname{cosec} 60^{\circ}=\sec 30^{\circ}=\frac{2}{\sqrt{3}}, \\ \cot 45^{\circ}=\tan 45^{\circ}=1 \end{array}\right]} \\ \end{aligned}\)
\(=\frac{3 \sqrt{3}-4}{4+3 \sqrt{3}}=\frac{3 \sqrt{3}-4}{4+3 \sqrt{3}} \times \frac{4-3 \sqrt{3}}{4-3 \sqrt{3}}\)
[multiplying numerator and denominator by the conjugate of 4 + 3\(\sqrt3\), i.e. 4 - 3\(\sqrt3\)]
\(\begin{aligned} & =\frac{12 \sqrt{3}-27-16+12 \sqrt{3}}{(4)^2-(3 \sqrt{3})^2}\left[\because(a+b)(a-b)=a^2-b^2\right] \\ \end{aligned}\)
\(\begin{aligned} =\frac{24 \sqrt{3}-43}{16-27}=\frac{-(43-24 \sqrt{3})}{-11}=\frac{43-24 \sqrt{3}}{11} \end{aligned}\)

\(\frac { { tan }^{ 2 }60°+{ 4sin }^{ 2 }45°+{ 3sec }^{ 2 }30°+{ 5cos }^{ 2 }90° }{ cosec30°+sec60°-{ cot }^{ 2 }30° } \)
=\(\frac { { \left( 3 \right) }^{ 2 }+4\times { \left( \frac { 1 }{ \sqrt { 2 } } \right) }^{ 2 }+3\times { \left( \frac { 2 }{ \sqrt { 3 } } \right) }^{ 2 }+5\times 0 }{ 2+2-{ \left( \sqrt { 3 } \right) }^{ 2 } } \)
=\(\frac{3+2+4}{1}\)=9

\(\cos A=\frac{5}{13}\)
\(\tan A=\frac{12}{5},\)
\(\\ \cot A=\frac{5}{12}, \ \)
\(\\sec A=\frac{13}{5},\)
\(\operatorname{cosec} A=\frac{13}{12}\)

\(\cos A=\frac{4}{5}\)
\(\tan A=\frac{3}{4}\)
\(\cot A=\frac{4}{3}\)
\(\sec A=\frac{5}{4}\)
\(\operatorname{cosec} A=\frac{5}{3}\) 

- 2

\( \frac{3}{2}\) 

\(\frac{29 \sqrt{3}+1}{4 \sqrt{3}}\)

\(\frac{3(1-\sqrt{3})}{2}\)

x = 1

15^o

4

A = 60° , B = 30°

Put \(\cos 60^{\circ}=\sin 30^{\circ}=\frac{1}{2}\)
\(\tan 60^{\circ}=\sqrt{3}, \cot 45^{\circ}=1, \sec 30^{\circ}=\frac{2}{\sqrt{3}}\) in LHS and simplify.

We have, 5 sin² 45° - sec 60° cot²30°
\(\begin{aligned} =\left[5 \times\left(\frac{1}{\sqrt{2}}\right)^2\right]-\left(2 \times(\sqrt{3})^2\right) \\ \end{aligned}\)
\(\begin{aligned} =\left(5 \times \frac{1}{2}\right)-(2 \times 3)=\frac{5}{2}-6=\frac{5-12}{2}=-\frac{7}{2} \end{aligned}\)

\(\because\) We know that
\(\cos 45^{\circ}=\frac{1}{\sqrt{2}}, \sin 60^{\circ}=\frac{\sqrt{3}}{2}, \sec 30^{\circ}=\frac{2}{\sqrt{3}} \text { and } \operatorname{cosec} 30^{\circ}=2\)
Now, \(\frac{\cos 45^{\circ}+\sin 60^{\circ}}{\operatorname{sec} 30^{\circ}+\operatorname{cosec} 30^{\circ}}=\frac{\frac{1}{\sqrt{2}}+\frac{\sqrt{3}}{2}}{\frac{2}{\sqrt{3}}+2}=\frac{\frac{2+\sqrt{6}}{2 \sqrt{2}}}{\frac{2+2 \sqrt{3}}{\sqrt{3}}}\)
\(=\frac{(2+\sqrt{6}) \sqrt{3}}{2 \sqrt{2}(2+2 \sqrt{3})}=\frac{2 \sqrt{3}+\sqrt{18}}{4 \sqrt{2}+4 \sqrt{6}}\)
\(=\frac{2 \sqrt{3}+3 \sqrt{2}}{4 \sqrt{2}+4 \sqrt{6}}=\frac{2 \sqrt{3}+3 \sqrt{2}}{4 \sqrt{2}+4 \sqrt{6}} \times \frac{4 \sqrt{2}-4 \sqrt{6}}{4 \sqrt{2}-4 \sqrt{6}}\)
\(=\frac{8 \sqrt{6}-8 \sqrt{18}+12 \times 2-12 \sqrt{12}}{32-96}\)
\(=\frac{8 \sqrt{6}-24 \sqrt{2}+24-24 \sqrt{3}}{-64}=\frac{-\sqrt{6}+3 \sqrt{2}-3+3 \sqrt{3}}{8}\)
\(=\frac{+\sqrt{2}(-\sqrt{3}+3)+\sqrt{3}(-\sqrt{3}+3)}{8}=\frac{(-\sqrt{3}+3)(\sqrt{2}+\sqrt{3})}{8}\)
\(\therefore \frac{\cos 45^{\circ}+\sin 60^{\circ}}{\sec 30^{\circ}+\operatorname{cosec} 30^{\circ}}=\frac{(-\sqrt{3}+3)(\sqrt{2}+\sqrt{3})}{8}\)

We have to evaluate
\(\frac{\sec ^2 45^{\circ}-\tan ^2 45^{\circ}}{\sin ^2 45^{\circ}}\)
We know that sec 45° = \(\sqrt{2}\), tan 45° = 1
and sin 45°= \(\frac{1}{\sqrt{2}}\)
Now, on putting all the values, we get
\(\Rightarrow \quad \frac{(\sqrt{2})^2-(1)^2}{\left(\frac{1}{\sqrt{2}}\right)^2} \Rightarrow \frac{2-1}{\frac{1}{2}}=\frac{1}{\frac{1}{2}}=2\)

To find \(\frac{5 \tan 60^{\circ}}{\left(\sin ^2 60^{\circ}+\cos ^2 60^{\circ}\right) \tan 30^{\circ}}\)         ...(i)
We have, \(\tan 60^{\circ}=\sqrt{3}, \tan 30^{\circ}=\frac{1}{\sqrt{3}}, \sin 60^{\circ}=\frac{\sqrt{3}}{2}\)
and \(\cos 60^{\circ}=\frac{1}{2}\)
On substituting the values of tan60°, tan 30° and sin60° and cos60° in Eq. (i), we get
\(\begin{gathered} \frac{5 \times \sqrt{3}}{\left[\left(\frac{\sqrt{3}}{2}\right)^2+\left(\frac{1}{2}\right)^2\right] \times \frac{1}{\sqrt{3}}} \\ \end{gathered}\)
\(\begin{gathered} =\frac{5 \sqrt{3}}{\left(\frac{3}{4}+\frac{1}{4}\right) \times \frac{1}{\sqrt{3}}} \end{gathered}\)
\(\Rightarrow \quad 5 \sqrt{3} \times \sqrt{3}=15\)
Therefore, \(\frac{5 \tan 60^{\circ}}{\left(\sin ^2 60^{\circ}+\cos ^2 60^{\circ}\right) \tan 30^{\circ}}=15\)