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,    \(\sin \mathrm{Q}=\frac{7}{25} \text { and } \cos \mathrm{Q}=\frac{24}{25}\)

In  \(\triangle\)ACB, we have
\(\mathrm{AC}=\sqrt{\mathrm{AB}^{2}-\mathrm{BC}^{2}}=\sqrt{(29)^{2}-(21)^{2}}\)
\(=\sqrt{(29-21)(29+21)}=\sqrt{(8)(50)}=\sqrt{400}=20 \text { units }\)
So,   \(\sin \theta=\frac{\mathrm{AC}}{\mathrm{AB}}=\frac{20}{29}, \cos \theta=\frac{\mathrm{BC}}{\mathrm{AB}}=\frac{21}{29}\)
Now, (i) cos²\(\theta\) + sin²\(\theta\)  \(=\left(\frac{20}{29}\right)^{2}+\left(\frac{21}{29}\right)^{2}=\frac{20^{2}+21^{2}}{29^{2}}=\frac{400+441}{841}=1\) 
and (ii) cos²\(\theta\) + sin²\(\theta\)  \(=\left(\frac{21}{29}\right)^{2}-\left(\frac{20}{29}\right)^{2}=\frac{(20+20)(21-20)}{29^{2}}=\frac{41}{841}\) 

Given PQ = 3 cm and PR = 6 cm.
Therefore,  \(\begin{aligned} \frac{\mathrm{PQ}}{\mathrm{PR}} & =\sin \mathrm{R} \\ \end{aligned}\)
or  \(​​​​\begin{aligned} \sin \mathrm{R} & =\frac{3}{6}=\frac{1}{2} \\ \end{aligned}\)
So, \(\begin{aligned} \angle \mathrm{PRQ} & =30^{\circ} \\ \end{aligned}\)
and therefore, \(\begin{aligned} \angle \mathrm{QPR} & =60^{\circ} \end{aligned}\)
You may note that if one of the sides and any other part (either an acute angle or any side) of a right triangle is known, the remaining sides and angles of the triangle can be determined.

To find the length of the side BC, we will choose the trigonometric ratio involving BC and the given side AB. Since BC is the side adjacent to angle C and AB is the side opposite to angle C, therefore

\(\begin{aligned} & \frac{\mathrm{AB}}{\mathrm{BC}}=\tan \mathrm{C} \\ \end{aligned}\)
i.e.,   \(\begin{aligned} & \frac{5}{\mathrm{BC}}=\tan 30^{\circ}=\frac{1}{\sqrt{3}} \\ \end{aligned}\)
which gives \(\begin{aligned} & \mathrm{BC}=5 \sqrt{3} \mathrm{~cm} \end{aligned}\)
To find the length of the side AC, we consider
\(\begin{aligned} \sin 30^{\circ} & =\frac{\mathrm{AB}}{\mathrm{AC}} \\ \end{aligned}\)
i.e., \(\begin{aligned} \frac{1}{2} & =\frac{5}{\mathrm{AC}} \\ \end{aligned}\)
i.e., AC = 10 cm
Note that alternatively we could have used Pythagoras theorem to determine the third side in the example above,
i.e., \(\mathrm{AC}=\sqrt{\mathrm{AB}^2+\mathrm{BC}^2}=\sqrt{5^2+(5 \sqrt{3})^2} \mathrm{~cm}=10 \mathrm{~cm} .\)

Let us first draw a right \(\Delta\)ABC

Now, we know that tan \(\mathrm{A}=\frac{\mathrm{BC}}{\mathrm{AB}}=\frac{4}{3}\)
Therefore, if BC = 4k, then AB = 3k, where k is a positive number.
Now, by using the Pythagoras Theorem, we have
AC² = AB² + BC² = (4k)² + (3k)² = 25k²
So, AC = 5k
Now, we can write all the trigonometric ratios using their definitions.
\(\begin{aligned} & \sin \mathrm{A}=\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{4 k}{5 k}=\frac{4}{5} \\ \end{aligned}\)
\(\begin{aligned} & \cos \mathrm{A}=\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{3 k}{5 k}=\frac{3}{5} \end{aligned}\)
Therefore, cot \(\mathrm{A}=\frac{1}{\tan \mathrm{A}}=\frac{3}{4}, \operatorname{cosec} \mathrm{A}=\frac{1}{\sin \mathrm{A}}=\frac{5}{4} \text { and } \sec \mathrm{A}=\frac{1}{\cos \mathrm{A}}=\frac{5}{3} .\)

Since we will apply the identity involving sec \(\theta\) and tan \(\theta\), let us first convert the LHS (of the identity we need to prove) in terms of sec \(\theta\) and tan \(\theta\) by dividing numerator and denominator by cos \(\theta\).
\(\begin{aligned} \text { LHS } & =\frac{\sin \theta-\cos \theta+1}{\sin \theta+\cos \theta-1}=\frac{\tan \theta-1+\sec \theta}{\tan \theta+1-\sec \theta} \\ \end{aligned}\)
\(\begin{aligned} & =\frac{(\tan \theta+\sec \theta)-1}{(\tan \theta-\sec \theta)+1}=\frac{\{(\tan \theta+\sec \theta)-1\}(\tan \theta-\sec \theta)}{\{(\tan \theta-\sec \theta)+1\}(\tan \theta-\sec \theta)} \\ \end{aligned}\)
\(\begin{aligned} & =\frac{\left(\tan ^2 \theta-\sec ^2 \theta\right)-(\tan \theta-\sec \theta)}{\{\tan \theta-\sec \theta+1\}(\tan \theta-\sec \theta)} \\ \end{aligned}\)
\(\begin{aligned} & =\frac{-1-\tan \theta+\sec \theta}{(\tan \theta-\sec \theta+1)(\tan \theta-\sec \theta)} \end{aligned}\)
\(=\frac{-1}{\tan \theta-\sec \theta}=\frac{1}{\sec \theta-\tan \theta}\)
which is the RHS of the identity, we are required to prove.

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°.

LHS = \(\frac { \cos { A } -\sin { A } +1 }{ \cos { A } +\sin { A } -1 } \)
On dividing numerator and denominator by sin A, we get
\(=\frac { \frac { \cos { A } }{ \sin { A } } -\frac { \sin { A } }{ \sin { A } } +\frac { 1 }{ \sin { A } } }{ \frac { \cos { A } }{ \sin { A } } +\frac { \sin { A } }{ \sin { A } } -\frac { 1 }{ \sin { A } } } =\frac { \cot { A } -1+cosecA }{ \cot { A } +1-1cosecA } \) \(\left[ \because \cot { A } =\frac { \cos { \theta } }{ \sin { \theta } } and\frac { 1 }{ \sin { \theta } } =cosec \theta \right] \)
\(=\frac { \cot { A } +cosecA-1 }{ \cot { A } +1-1cosecA } \)
\(=\frac { (\cot { A } +cosecA)-({ cosec }^{ 2 }A-\cot ^{ 2 }{ A } ) }{ \cot { A } +1-1cosecA } \left[ \because 1={ cosec }^{ 2 }A-\cot ^{ 2 }{ A } \right] \)
\(=\frac { (\cot { A } +cosecA)-\left[ (\cot { A } +cosecA)(\cot { A } -cosecA) \right] }{ \cot { A } +1-1cosecA } \) \(\left[ \because \quad { a }^{ 2 }-{ b }^{ 2 }=(a+b)(a-b) \right] \)
\(=\frac { (\cot { A } +cosecA)-\left[ 1-(cosecA-\cot { A } ) \right] }{ \cot { A } +1-1cosecA } \)
\(\quad =\frac { (\cot { A } +cosecA)-\left[ 1-cosecA+\cot { A } \right] }{ \cot { A } +1-1cosecA } \)
\(=cosecA+\cot { A } =RHS\)
Hence proved.

LHS = \(\frac { \cos { A } }{ 1+\sin { A } } +\frac { 1+\sin { A } }{ \cos { A } } =\frac { \cos ^{ 2 }{ A } +{ (1+\sin { A } ) }^{ 2 } }{ (1+\sin { A } )\cos { A } } \)
\(=\frac { \cos ^{ 2 }{ A } +\sin ^{ 2 }{ A } +1+2\sin { A } }{ (1+\sin { A } )\cos { A } } \) \(\left[ \because { (a+b) }^{ 2 }={ a }^{ 2 }+{ b }^{ 2 }+2ab \right] \)
\(=\frac { 1+1+2\sin { A } }{ (1+\sin { A } )\cos { A } } \left[ \because \sin ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta=1 } \right] \)
\(=\frac { 2+2\sin { A } }{ (1+\sin { A } )\cos { A } } =\frac { 2(1+\sin { A } ) }{ (1+\sin { A } )\cos { A } } \)
\(=\frac { 2 }{ \cos { A } } =2\sec { A } \left[ \because \frac { 1 }{ \cos { \theta } }=\sec { \theta } \right] \)
= RHS
Hence proved.

LHS=\({ (cosec\theta -\cot { \theta } ) }^{ 2 }={ \left( \frac { 1 }{ \sin { \theta } } -\frac { \cos { \theta } }{ \sin { \theta } } \right) }^{ 2 }\) \(\left[ \because cosec \quad A =\frac { 1 }{ \sin { A } } ,\cot { A } =\frac { \cos { A } }{ \sin { A } } \right] \)
\(={ \left( \frac { 1-\cos { \theta } }{ \sin { \theta } } \right) }^{ 2 }=\frac { { (1-\cos { \theta } ) }^{ 2 } }{ \sin ^{ 2 }{ \theta } } =\frac { { (1-\cos { \theta } ) }^{ 2 } }{ 1-\cos ^{ 2 }{ \theta } } \left[ \because \sin ^{ 2 }{ A } =1-\cos ^{ 2 }{ A} \right] \)
\(=\frac { (1-\cos { \theta } )(1-\cos { \theta } ) }{ (1+\cos { \theta } )(1-\cos { \theta } ) } \left[ \because \quad { a }^{ 2 }-{ b }^{ 2 }=(a+b)(a-b) \right] \)
\(=\frac { 1-\cos { \theta } }{ 1+\cos { \theta } } =RHS\)
Hence proved.

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

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