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°.

We have, sec 5A = cosec (A + 30⁰)
\(\Rightarrow\) sec 5A = sec[90⁰- (A + 30⁰)]  \([\because sec({ 90 }^{ 0 }-\theta )=cosec\theta ]\)
\(\Rightarrow\) sec 5A = sec(60⁰ - A)
\(\Rightarrow\) 5A = 60⁰ - A  [\(\therefore\) 5A and (60⁰- A) are acute angles]
\(\Rightarrow\) 6A = 60⁰
\(\therefore\) A = 10⁰ 

Statue=CD = 1.6 m
h = height of pedestal=BC
A is point on earth
In \(\triangle\)ABD, \(\cot { { 60 }^{ 0 } } =\frac { AB }{ BD } \)
\(\Rightarrow \quad \frac { 1 }{ \sqrt { 3 } } =\frac { AB }{ h+1.6 } \)
\(\Rightarrow \quad AB=\frac { h+1.6 }{ \sqrt { 3 } } \quad ...(i)\)
\(\triangle\)ABC, \(\frac { AB }{ BC } =\cot { { 45 }^{ 0 } } \)
\(\Rightarrow \quad \frac { 1 }{ \sqrt { 3 } } =\frac { AB }{ h+1.6 } \)
\(\Rightarrow \quad \frac { 1 }{ \sqrt { 3 } } =\frac { AB }{ h+1.6 } \)
\(\Rightarrow \quad AB=h\)
From (i) and (ii), we get
\(\Rightarrow \quad h=\frac { h+1.6 }{ \sqrt { 3 } } \)
Height of pedestal = h =2.2m.

\(\cos { { 50 }^{ ° } } =\cos { \left( { 90 }^{ ° }-{ 40 }^{ ° } \right) } =\sin { { 40 }^{ ° } } \)
\({ cosec }^{ 2 }{ 59 }^{ ° }={ cosec }^{ 2 }\left( { 90 }^{ ° }-{ 31 }^{ ° } \right) =\sec ^{ 2 }{ { 31 }^{ ° } } \)
and \(\tan { { 78 }^{ ° } } =\tan { \left( { 90 }^{ ° }-{ 12 }^{ ° } \right) } =\cot { { 12 }^{ ° } } \)
Hence, \(\frac { \cos { { 50 }^{ ° } } }{ 2\sin { { 40 }^{ ° } } } +\frac { 4\left( { cosec }^{ 2 }{ 59 }^{ ° }-\tan ^{ 2 }{ { 31 }^{ ° } } \right) }{ 3\tan ^{ 2 }{ { 45 }^{ ° } } } -\frac { 2 }{ 3 } \tan { { 12 }^{ ° } } \tan { { 78 }^{ ° } } .\sin { { 90 }^{ ° } } \)
\(\frac { \sin { { 40 }^{ ° } } }{ 2\sin { { 40 }^{ ° } } } +\frac { 4\left( \sec ^{ 2 }{ { 31 }^{ ° } } -\tan ^{ 2 }{ { 31 }^{ ° } } \right) }{ 3\times { \left( 1 \right) }^{ 2 } } -\frac { 2 }{ 3 } \tan { { 12 }^{ ° } } \cot { { 12 }^{ ° } } \times 1\)
\(=\frac{1}{2}+\frac{4}{3}-\frac{2}{3}=\frac{7}{6}\)

In \(\triangle\) OPQ, we have
OQ² = OP² + PQ²
i.e., (1 + PQ)² = OP² + PQ²
i.e., 1 + PQ² + 2PQ = OP² + PQ²
i.e., 1 + 2PQ = 7²
i.e., PQ = 24 cm and OQ = 1 + PQ = 25 cm
So,    \(\sin \mathrm{Q}=\frac{7}{25} \text { and } \cos \mathrm{Q}=\frac{24}{25}\)

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

\(\frac {1}{2}\)

False, let A = 60⁰ and B = 30⁰
Then, sin (A + B) = sin (60⁰ + 30⁰) = sin 90⁰ = 1
and sin A+sin B=sin 60⁰ + sin 30⁰
\(=\frac { \sqrt { 3 } }{ 2 } +\frac { 1 }{ 2 } =\frac { \sqrt { 3 } +1 }{ 2 } \)
So, sin (A + B) \(\ne\)  sin A + sin B

LHS = \(\frac { \sin { \theta } +2\sin ^{ 3 }{ \theta } }{ 2\cos ^{ 3 }{ \theta } -\cos { \theta } } =\frac { \sin { \theta } (1+2\sin ^{ 2 }{ \theta } ) }{ \cos { \theta } (2\cos ^{2 }{ \theta } -1) } \)
\(=\frac { \sin { \theta } [1-2(1-\cos ^{ 2 }{ \theta } )] }{ \cos { \theta } (2\cos ^{ 2 }{ \theta } -1) } \quad \left[ \because \sin ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta } =1\Rightarrow \sin ^{ 2 }{ \theta } =1-\cos ^{ 2 }{ \theta } \right] \)
\(=\frac { \sin { \theta } (1-2+2\cos ^{ 2 }{ \theta } ) }{ \cos { \theta } (2\cos ^{ 2 }{ \theta } -1) } \)
\(=\frac { \sin { \theta } (2\cos ^{ 2 }{ \theta } -1) }{ \cos { \theta } (2\cos ^{ 2 }{ \theta } -1) } =\frac { \sin { \theta } }{ \cos { \theta } } =\tan { \theta } =RHS\quad \left[ \because \tan { \theta } =\frac { \sin { \theta } }{ \cos { \theta } } \right] \)
Hence proved.

Use Pythagoras theorem to find the hypotenuse.
5 cm

Given, \(A+B={ 90 }^{ 0 }\quad \Rightarrow \quad B={ 90 }^{ 0 }-A\)
Now, LHS=\(\sqrt { \frac { \tan { A } \tan { B } +\tan { A } \cot { B } }{ \sin { A } \sec { B } } -\frac { \sin ^{ 2 }{ A } }{ \cos ^{ 2 }{ A } } } \)
\(=\sqrt { \frac { \tan { A } \tan { ({ 90 }^{ 0 }-A) } +\tan { A } \cot { ({ 90 }^{ 0 }-A) } }{ \sin { A } \sec { ({ 90 }^{ 0 }-A) } } -\frac { \sin ^{ 2 }{ ({ 90 }^{ 0 }-A) } }{ \cos ^{ 2 }{ A } } } \)
\(=\sqrt { \frac { \tan { A } \tan { A } +\tan { A } \tan { A } }{ \sin { A } cosecA } -\frac { \cos ^{ 2 }{ A } }{ \cos ^{ 2 }{ A } } } \)
\(=\sqrt { 1+\tan ^{ 2 }{ A } -1 } \)
\(=\sqrt { \tan ^{ 2 }{ A } } =\tan { A } =RHS\) 

\(LHS=-1+\frac { \sin { A } \sin { \left( { 90 }^{ ° }-A \right) } }{ \cot { \left( { 90 }^{ ° }-A \right) } } \)
\(\left[ \because \quad \sin { \left( { 90 }^{ ° }-\theta \right) =\cos { \theta } } \right] \)
\(\left[ \because \quad \cot { \left( { 90 }^{ ° }-\theta \right) } =\tan { \theta } \right] \)
\(=-1+\frac { \sin { A } \cos { A } }{ \tan { A } } \)
= - 1 + sin A cos A x cot A
\(\left[ \because \quad \cot { \theta } =\frac { \sin { \theta } }{ \cos { \theta } } \right] \)
\(\left[ \because \quad \sin ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta } =1 \right] \)
\(=-1+\sin { A } \cos { A } \times \frac { \cos { A } }{ \sin { A } } \)
= - 1 + cos² A = - (1 - cos²A)
= - sin²A = RHS

\(L H S=\frac{\cot A-\cos A}{\cot A+\cos A}=\frac{\frac{\cos A}{\sin A}-\cos A}{\frac{\cos A}{\sin A}+\cos A}\)
\(=\frac{\cos A\left(\frac{1}{\sin A}-1\right)}{\cos A\left(\frac{1}{\sin A}+1\right)}=\frac{\left(\frac{1}{\sin A}-1\right)}{\left(\frac{1}{\sin A}+1\right)}=\frac{\operatorname{cosec} A-1}{\operatorname{cosec} A+1}=\operatorname{RHS}\)

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

Given, 22 cosA - 3s inA = 20 sinA
\(\Rightarrow \quad \frac{22}{20} \cot A-\frac{3}{20}=1 \Rightarrow \cot A=\frac{23}{22} \Rightarrow \tan A=\frac{22}{23}\)
Now, put in given expression and simplify
\(2 \times\left(\frac{22}{23}\right)^{2}\) 

(i) (a): We have, AB = 9 m, BC = 3\(\sqrt{3}\) m
In  \(\Delta\)ABC, we have
\(\tan A=\frac{B C}{A B}=\frac{3 \sqrt{3}}{9}=\frac{1}{\sqrt{3}} \)
\(\Rightarrow \tan A=\tan 30^{\circ} \Rightarrow \angle A=30^{\circ}\)
(ii) (c): Similarly, \(\tan C=\frac{A B}{B C}=\frac{9}{3 \sqrt{3}}=\sqrt{3}\)
\(\Rightarrow \tan C=\tan 60^{\circ} \Rightarrow \angle C=60^{\circ}\)
(iii) (d): Since \(\sin A=\frac{B C}{A C} \Rightarrow \sin 30^{\circ}=\frac{B C}{A C}\)
\(\Rightarrow \frac{1}{2}=\frac{3 \sqrt{3}}{A C} \Rightarrow A C=6 \sqrt{3} \mathrm{~m}\)
(iv) (b) : \(\because \angle A=30^{\circ}\) [From (1)]
\(\therefore \quad \cos 2 A=\cos \left(2 \times 30^{\circ}\right)=\cos 60^{\circ}=\frac{1}{2}\) 
(v) (b): \(\because \angle C=60^{\circ}\) [Using (2)]
\(\therefore \quad \sin \left(\frac{C}{2}\right)=\sin \left(\frac{60^{\circ}}{2}\right)=\sin 30^{\circ}=\frac{1}{2}\)

\(\because \Delta\)PQR is a right angled triangle.
\(\therefore\) PR² + RQ² = PQ²
\(\Rightarrow P R^{2}=(13)^{2}-(12)^{2}=25 \Rightarrow P R=5 \mathrm{~cm}\)
(i) (c) : \(\cos \theta=\frac{Q R}{P Q}=\frac{12}{13} \)
(ii) (c) : \(\sec \theta=\frac{1}{\cos \theta}=\frac{13}{12} \)
(iii) (c) : \(\tan \theta=\frac{P R}{R Q}=\frac{5}{12}\)
\(\therefore \frac{\tan \theta}{1+\tan ^{2} \theta}=\frac{\frac{5}{12}}{1+\frac{25}{144}}=\frac{\frac{5}{12}}{\frac{169}{144}}=\frac{60}{169}\)
(iv) (a): \(\cot \theta=\frac{1}{\tan \theta}=\frac{12}{5}\)    [Using (1)]
\(\operatorname{cosec} \theta=\frac{P Q}{P R}=\frac{13}{5} \)
\(\therefore \quad \cot ^{2} \theta-\operatorname{cosec}^{2} \theta=\frac{144}{25}-\frac{169}{25}=-1\)
(v) (b): \(\sin ^{2} \theta+\cos ^{2} \theta=1\) (Using identity)

We have, AB = BC = 6\(\sqrt{2}\) m and AC=12m .
(i) (d):\(\because\)  Dis mid point of AC.
\(\therefore\) AD=DC=6m
Now, AB² = BD² + AD² (\(\therefore\) \(\Delta\)ABD is a right triangle)
\(\Rightarrow B D^{2}=(6 \sqrt{2})^{2}-6^{2}=72-36=36 \)
\(\Rightarrow B D=6 \mathrm{~m}\)
(ii) (c) : \(\operatorname{In} \Delta A B D, \sin A=\frac{B D}{A B}=\frac{6}{6 \sqrt{2}}=\frac{1}{\sqrt{2}}\)
\(\Rightarrow \sin A=\sin 45^{\circ} \Rightarrow \angle A=45^{\circ}\)
(iii) (c) : \(\operatorname{In} \Delta B D C, \tan C=\frac{B D}{D C}=\frac{6}{6}\)
\(\Rightarrow \tan C=1=\tan 45^{\circ} \Rightarrow \angle C=45^{\circ}\)
(iv) (d) : \(\sin A=\frac{1}{\sqrt{2}}, \cos C=\cos 45^{\circ}=\frac{1}{\sqrt{2}}\)
\(\therefore \quad \sin A+\cos C=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\frac{2}{\sqrt{2}}=\sqrt{2}\)
(v) (c): \((v) \quad(c): \tan C=1, \tan A=\tan 45^{\circ}=1\)
\(\Rightarrow \tan ^{2} C+\tan ^{2} A=1+1=2\)