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): \((v) \quad(c): \tan C=1, \tan A=\tan 45^{\circ}=1\)
\(\Rightarrow \tan ^{2} C+\tan ^{2} A=1+1=2\)
 

We have, KL = 4 cm, ML = 4\(\sqrt{3}\)m, KM = 8 cm
(i) (a): \(\tan M=\frac{K L}{L M}=\frac{4}{4 \sqrt{3}}=\frac{1}{\sqrt{3}}\)
\(\Rightarrow \tan M=\tan 30^{\circ} \Rightarrow \angle M=30^{\circ}\)
(ii) (c) : \(\tan K=\frac{M L}{K L}=\frac{4 \sqrt{3}}{4}=\sqrt{3}=\tan 60^{\circ}\)
\(\Rightarrow \angle K=60^{\circ}\)
(iii) (b)
(iv) (c)
(v) (a) : \(\frac{\tan ^{2} 45^{\circ}-1}{\tan ^{2} 45^{\circ}+1}=\frac{(1)^{2}-1}{1^{2}+1}=\frac{0}{2}=0\)

\(\because \Delta\)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)

(i) (d) : \(\text { In } \Delta A P Q, \tan \theta=\frac{A Q}{P Q}=\frac{1.2}{1.6}=\frac{3}{4}\)
(ii) (d) : \(\text { In } \Delta P B Q, \cot B=\frac{Q B}{P Q}=\frac{3}{1.6}=\frac{15}{8}\) ...(i)
(iii) (c): \(\text { In } \Delta A P Q, \tan A=\frac{P Q}{A Q}=\frac{1.6}{1.2}=\frac{4}{3}\) ...(ii)
(iv) (d): We have, tan²A + 1 = sec²A
\(\Rightarrow \sec A =\sqrt{\left(\frac{4}{3}\right)^{2}+1} \)
\(=\sqrt{\frac{16}{9}+1}=\sqrt{\frac{25}{9}}=\frac{5}{3}\)
(v) (a): Since \(\operatorname{cosec} B=\sqrt{\cot ^{2} B+1}\)
\(\begin{array}{l} =\sqrt{\left(\frac{15}{8}\right)^{2}+1} \\ =\frac{17}{8} \end{array}\)

(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)]
\(\therefore \quad \sin \left(\frac{C}{2}\right)=\sin \left(\frac{60^{\circ}}{2}\right)=\sin 30^{\circ}=\frac{1}{2}\)