Solution

(i) $\sin 60^{\circ} \cos 30^{\circ}+\sin 30^{\circ} \cos 60^{\circ}$

$=(\dfrac{\sqrt{3}}{2})(\dfrac{\sqrt{3}}{2})+(\dfrac{1}{2})(\dfrac{1}{2})$

$=\dfrac{3}{4}+\dfrac{1}{4}=\dfrac{4}{4}=1$

(ii) $2 \tan ^{2} 45^{\circ}+\cos ^{2} 30^{\circ}-\sin ^{2} 60^{\circ}$

$=2(1)^{2}+(\dfrac{\sqrt{3}}{2})^{2}-(\dfrac{\sqrt{3}}{2})^{2}$

$=2+\dfrac{3}{4}-\dfrac{3}{4}=2$

(iii) $\dfrac{\cos 45^{\circ}}{\sec 30^{\circ}+cosec 30^{\circ}}$

$ \begin{aligned} & =\dfrac{\dfrac{1}{\sqrt{2}}}{\dfrac{2}{\sqrt{3}}+2}=\dfrac{\dfrac{1}{\sqrt{2}}}{\dfrac{2+2 \sqrt{3}}{\sqrt{3}}} \\ & =\dfrac{\sqrt{3}}{\sqrt{2}(2+2 \sqrt{3})}=\dfrac{\sqrt{3}}{2 \sqrt{2}+2 \sqrt{6}} \\ & =\dfrac{\sqrt{3}(2 \sqrt{6}-2 \sqrt{2})}{(2 \sqrt{6}+2 \sqrt{2})(2 \sqrt{6}-2 \sqrt{2})} \end{aligned} $

$=\dfrac{2 \sqrt{3}(\sqrt{6}-\sqrt{2})}{(2 \sqrt{6})^{2}-(2 \sqrt{2})^{2}}=\dfrac{2 \sqrt{3}(\sqrt{6}-\sqrt{2})}{24-8}=\dfrac{2 \sqrt{3}(\sqrt{6}-\sqrt{2})}{16}$

$=\dfrac{\sqrt{18}-\sqrt{6}}{8}=\dfrac{3 \sqrt{2}-\sqrt{6}}{8}$

$\dfrac{\sin 30^{\circ}+\tan 45^{\circ}-cosec 60^{\circ}}{\sec 30^{\circ}+\cos 60^{\circ}+\cot 45^{\circ}}$

$=\dfrac{\dfrac{1}{2}+1-\dfrac{2}{\sqrt{3}}}{\dfrac{2}{\sqrt{3}}+\dfrac{1}{2}+1}=\dfrac{\dfrac{3}{2}-\dfrac{2}{\sqrt{3}}}{\dfrac{3}{2}+\dfrac{2}{\sqrt{3}}}$

$=\dfrac{\dfrac{3 \sqrt{3}-4}{2 \sqrt{3}}}{\dfrac{3 \sqrt{3}+4}{2 \sqrt{3}}}=\dfrac{(3 \sqrt{3}-4)}{(3 \sqrt{3}+4)}$

$=\dfrac{(3 \sqrt{3}-4)(3 \sqrt{3}-4)}{(3 \sqrt{3}+4)(3 \sqrt{3}-4)}=\dfrac{(3 \sqrt{3}-4)^{2}}{(3 \sqrt{3})^{2}-(4)^{2}}$

$=\dfrac{27+16-24 \sqrt{3}}{27-16}=\dfrac{43-24 \sqrt{3}}{11}$

(v) $\dfrac{5 \cos ^{2} 60^{\circ}+4 \sec ^{2} 30^{\circ}-\tan ^{2} 45^{\circ}}{\sin ^{2} 30^{\circ}+\cos ^{2} 30^{\circ}}$

$ \begin{aligned} & =\dfrac{5(\dfrac{1}{2})^{2}+4(\dfrac{2}{\sqrt{3}})^{2}-(1)^{2}}{(\dfrac{1}{2})^{2}+(\dfrac{\sqrt{3}}{2})^{2}} \\ & =\dfrac{5(\dfrac{1}{4})+(\dfrac{16}{3})-1}{\dfrac{1}{4}+\dfrac{3}{4}} \\ & =\dfrac{\dfrac{15+64-12}{12}}{\dfrac{4}{4}}=\dfrac{67}{12} \end{aligned} $

Solution

(i) $\dfrac{2 \tan 30^{\circ}}{1+\tan ^{2} 30^{\circ}}$

$=\dfrac{2(\dfrac{1}{\sqrt{3}})}{1+(\dfrac{1}{\sqrt{3}})^{2}}=\dfrac{\dfrac{2}{\sqrt{3}}}{1+\dfrac{1}{3}}=\dfrac{\dfrac{2}{\sqrt{3}}}{\dfrac{4}{3}}$

$=\dfrac{6}{4 \sqrt{3}}=\dfrac{\sqrt{3}}{2}$

Out of the given alternatives, only

$ \sin 60^{\circ}=\dfrac{\sqrt{3}}{2} $

Hence, (A) is correct.

(ii) $\dfrac{1-\tan ^{2} 45^{\circ}}{1+\tan ^{2} 45^{\circ}}$

$=\dfrac{1-(1)^{2}}{1+(1)^{2}}=\dfrac{1-1}{1+1}=\dfrac{0}{2}=0$

Hence, (D) is correct.

(iii)Out of the given alternatives, only $A=0^{\circ}$ is correct.

As $\sin 2 A=\sin 0^{\circ}=0$

$2 \sin A=2 \sin 0^{\circ}=2(0)=0$

Hence, $(A)$ is correct.

(iv) $\dfrac{2 \tan 30^{\circ}}{1-\tan ^{2} 30^{\circ}}$

$=\dfrac{2(\dfrac{1}{\sqrt{3}})}{1-(\dfrac{1}{\sqrt{3}})^{2}}=\dfrac{\dfrac{2}{\sqrt{3}}}{1-\dfrac{1}{3}}=\dfrac{\dfrac{2}{\sqrt{3}}}{\dfrac{2}{3}}$

$=\sqrt{3}$

Out of the given alternatives, only $\tan 60^{\circ}=\sqrt{3}$

Hence, (C) is correct.

Solution

$\tan (A+B)=\sqrt{3}$

$\Rightarrow \tan (A+B)=\tan 60$

$\Rightarrow A+B=60 \ldots$ (1)

$\tan (A-B)=\dfrac{1}{\sqrt{3}}$

$\Rightarrow \tan (A-B)=\tan 30$

$\Rightarrow A-B=30$..

On adding both equations, we obtain

$2 A=90$

$\Rightarrow A=45$

From equation (1), we obtain

$45+B=60$

$B=15$

Therefore, $\angle A=45^{\circ}$ and $\angle B=15^{\circ}$

Solution

(i) $\sin (A+B)=\sin A+\sin B$

Let $A=30^{\circ}$ and $B=60^{\circ}$

$\sin (A+B)=\sin (30^{\circ}+60^{\circ})$

$=\sin 90^{\circ}$

$=1$

$\sin A+\sin B=\sin 30^{\circ}+\sin 60^{\circ}$

$=\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}=\dfrac{1+\sqrt{3}}{2}$

Clearly, $\sin (A+B) \neq \sin A+\sin B$

Hence, the given statement is false.

(ii) The value of $\sin \theta$ increases as $\theta$ increases in the interval of $0^{\circ}<\theta<90^{\circ}$ as

$\sin 0^{\circ}=0$

$\sin 30^{\circ}=\dfrac{1}{2}=0.5$

$\sin 45^{\circ}=\dfrac{1}{\sqrt{2}}=0.707$

$\sin 60^{\circ}=\dfrac{\sqrt{3}}{2}=0.866$

$\sin 90^{\circ}=1$

Hence, the given statement is true.

(iii) $\cos 0^{\circ}=1$

$\cos 30^{\circ}=\dfrac{\sqrt{3}}{2}=0.866$

$\cos 45^{\circ}=\dfrac{1}{\sqrt{2}}=0.707$

$\cos 60^{\circ}=\dfrac{1}{2}=0.5$

$\cos 90^{\circ}=0$

It can be observed that the value of $\cos \theta$ does not increase in the interval of $0^{\circ}<\theta<90^{\circ}$.

Hence, the given statement is false.

(iv) $\sin \theta=\cos \theta$ for all values of $\theta$.

This is true when $\theta=45^{\circ}$

$ \begin{aligned} & \sin 45^{\circ}=\dfrac{1}{\sqrt{2}} \\ & \text{ As } \\ & \cos 45^{\circ}=\dfrac{1}{\sqrt{2}} \end{aligned} $

It is not true for all other values of $\theta$.

As $\sin 30^{\circ}=\dfrac{1}{2}$ and $\cos 30^{\circ}=\dfrac{\sqrt{3}}{2}$,

Hence, the given statement is false.

(v) $\cot A$ is not defined for $A=0^{\circ}$

$ \begin{aligned} & \text{ As } \cot A=\dfrac{\cos A}{\sin A} \\ & \cot 0^{\circ}=\dfrac{\cos 0^{\circ}}{\sin 0^{\circ}}=\dfrac{1}{0}=\text{ undefined } \end{aligned} $

Hence, the given statement is true.