First tan A \(=\sqrt{\sec ^{2} A-1}=\sqrt{\left(\frac{17}{8}\right)^{2} \mid-1}=\frac{15}{8}\)
Now, LHS \(=\frac{3-4 \sin ^{2} A}{4 \cos ^{2} A-3}=\frac{3 \sec ^{2} A-4 \tan ^{2} A}{4-3 \sec ^{2} A}\)
[dividing numerator and denominator by cos² A]
Now, put values of sec² A and tan ²A in LHS and RHS and hence verify the equality. 

Let AB be the tower of height h.
\(\angle AQB=45°\)
Now, in \(\Delta AQB\) ,

tan 45° =\(\frac{AB}{BQ}\)
1=\(\frac{h}{BQ}\)
BQ=h
In \(\Delta APB\),
tan 30° =\(\frac{AB}{PB}\)
x+h=h\(\sqrt{3}\)
\(\frac { 1 }{ \sqrt { 3 } } \)=\(\frac{h}{x+h}\)
x=h(\(\sqrt{3}\)-1)
Speed=\(\frac { h\left( \sqrt { 3 } -1 \right) }{ 12 } \)
Speed=Distance/Time
Time for remaining distance,
=\(\frac { h }{ \frac { h\left( \sqrt { 3 } -1 \right) }{ 12 } } \)
=\(\frac { 12\left( \sqrt { 3 } +1 \right) }{ 3-1 } \)
=\(\frac { 12 }{ 2 } \left( \sqrt { 3 } +1 \right) \)
= (6\(\sqrt{3}\)+1)
Time = 16·39minutes.

x = 3

\(A+B+C={ 180 }^{ 0 }\Rightarrow A+B={ 180 }^{ 0 }-C\Rightarrow A+B={ 90 }^{ 0 }\) 

\(\frac { \sqrt { 5 } }{ 3 } \)

LHS = (1+ tan A + cot A) (sin A - cos A)
\(=\left(1+\frac{\sin A}{\cos A}+\frac{\cos A}{\sin A}\right)(\sin A-\cos A)\)
\(=\left(\frac{\sin A \cos A+\sin ^{2} A+\cos ^{2} A}{\sin A \cos A}\right)(\sin A-\cos A)\)
\(=\frac{\sin ^{3} A-\cos ^{3} A}{\sin A \cos A}=\frac{\sin ^{2} A}{\cos A}-\frac{\cos ^{2} A}{\sin A}\)
= sin A tan A - cos A cot A = RHS
Hence proved.

ㄥAED = 60°, ㄥBEC = 30°
AD = Be = 3000ㄥ3 m

Let the speed of the aeroplane = x m/s
Then, AB = DC = 30 x x
= 30xm
ΔAED is right angled
\(tan\ 60^0={D\over DE}\)
\(\sqrt3={3000\sqrt3\over DE}\)
DE = 3000 m
∆BEC is right angled
\(tan\ 30^0={BC\over EC}\)
\({1\over \sqrt3}={3000\sqrt3\over DE+CD}\)
DE + CD = 3000 x 3
3000 + 30x = 9000
30x = 6000
x = 200 m/s
Speed of plane is 200 m/s.

Let P be the position of the pole and A and B be the oppositely fixed gates.
PA-PB = 7m
a-b = 7
a=7+b ...(i)
In \(\Delta PAB\),
AB²= AP² + BP²
(13)² = a² + b²
a² + b² = 169
(7 + b)² + b² = 169
49' + b² + 2(7)(b) + b² = 169
49 + 14b + 2b² - 169 = 0
2b² + 14b - 120 = 0
b² + 7b - 60 = 0
b² + 12b-5b-60 = 0
b(b + 12) - 5(b + 12) = 0
(b - 5)(b + 12) = 0

As b cannot be negative, so
b = 5
a = 7 + 5 = 12
Hence, PA = 12 m, PB = 5 m.

Let height of tower and distance BC = x m
In \(\triangle\)ABC,  \(\frac { h }{ x } =\tan { { 60 }^{ 0 } } \)
\(\Rightarrow \quad h=\sqrt { 3 } x\quad ...(i)\)
In \(\triangle\)ADC,
\(\frac { h }{ x+20 } =\tan { { 30 }^{ 0 } } =\frac { 1 }{ \sqrt { 3 } } \)
\(\Rightarrow \quad \sqrt { 3 } =x+20\quad \quad ...(ii)\)
Substituting the value of h from eg. (i) in eq. (ii), we get
3x = x + 20
\(\Rightarrow\) x= 10 m
\(\therefore\) AC= 30m.
Hence, height of tower is 10\(\sqrt3\) m and distance of tower form point A is 30 m.

Given :  \(x\sin { \theta } =y\cos { \theta } \)
\(\Rightarrow \quad x=\frac { y\cos { \theta } }{ \sin { \theta } } \)            ....(i)
and \(x\sin ^{ 3 }{ \theta } +y\cos ^{ 3 }{ \theta } =\sin { \theta } \cos { \theta } \)                  ...(ii)
Eliminating x from eqn.(i) amd eqn. (ii),
\(\frac { y\cos { \theta } }{ \sin { \theta } } \sin ^{ 3 }{ \theta } +y\cos ^{ 3 }{ \theta } =\sin { \theta } \cos { \theta } \)
\(\Rightarrow \quad y\cos { \theta } \sin ^{ 2 }{ \theta } +y\cos ^{ 3 }{ \theta } =\sin { \theta } \cos { \theta } \)
\(\Rightarrow \quad y\cos { \theta } \left[ \sin ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta } \right] =\sin { \theta } \cos { \theta } \)
\(\Rightarrow \quad y=\sin { \theta } \)                          ... (iii)
Substituting this value of y in eqn. (iii),
\(x=\cos { \theta } \)                            ... (iv)
Squaring and adding eqn. (iii) and eqn. (iv), we get
\({ x }^{ 2 }+{ y }^{ 2 }=\cos ^{ 2 }{ \theta } +\sin ^{ 2 }{ \theta } =1\)

Given, \(a\cos { \theta } -b\sin { \theta } =x\quad ...(i)\)
and \(a\cos { \theta } +b\sin { \theta } =y\quad ....(ii)\)
On squaring Eqs. (i) and (ii) and then adding, we get
\({ x }^{ 2 }+{ y }^{ 2 }={ \left( a\cos { \theta } -b\sin { \theta } \right) }^{ 2 }+{ \left( a\cos { \theta } +b\sin { \theta } \right) }^{ 2 }\)
\(\Rightarrow { x }^{ 2 }+{ y }^{ 2 }={ a }^{ 2 }\cos ^{ 2 }{ \theta } +{ b }^{ 2 }\sin ^{ 2 }{ \theta } -2ab\cos { \theta } \sin { \theta } +{ a }^{ 2 }\cos ^{ 2 }{ \theta } +{ b }^{ 2 }\sin ^{ 2 }{ \theta } +2ab\cos { \theta } \sin { \theta } \)\(\left[ \because { (a+b) }^{ 2 }={ a }^{ 2 }+{ b }^{ 2 }+2ab\quad and\quad { (a-b) }^{ 2 }={ a }^{ 2 }+{ b }^{ 2 }-2ab\quad \right] \)
\(\Rightarrow { x }^{ 2 }+{ y }^{ 2 }={ a }^{ 2 }\cos ^{ 2 }{ \theta } +{ b }^{ 2 }\sin ^{ 2 }{ \theta } +{ a }^{ 2 }\cos ^{ 2 }{ \theta } +{ b }^{ 2 }\sin ^{ 2 }{ \theta } \)
\(\Rightarrow { x }^{ 2 }+{ y }^{ 2 }={ a }^{ 2 }(\cos ^{ 2 }{ \theta } +\sin ^{ 2 }{ \theta } )+{ b }^{ 2 }(\cos ^{ 2 }{ \theta } +\sin ^{ 2 }{ \theta } )\)
\(\Rightarrow { x }^{ 2 }+{ y }^{ 2 }={ a }^{ 2 }+{ b }^{ 2 }\quad \quad \left[ \because \cos ^{ 2 }{ \theta } +\sin ^{ 2 }{ \theta } =1 \right] \)
Hence proved.

Value of \(\tan { \theta } \) when \(\theta\)  increases from 0⁰ to 90⁰ . 

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, 2 cos3 \(\theta=\sqrt{3}\)
\(\Rightarrow \quad \cos 3 \theta=\frac{\sqrt{3}}{2}\)
\(\Rightarrow\) cos3 \(\theta\) = cos30^o       \(\left[\because \cos 30^{\circ}=\frac{\sqrt{3}}{2}\right]\)
\(\Rightarrow\) 3\(\theta\)= 30^o as  3\(\theta\)= 30^o are acute angles.
\(\therefore\) \(\theta\) = 10^o

As,   \(\sqrt{3} \sin \theta=\cos \theta \Rightarrow \tan \theta=\frac{1}{\sqrt{3}}\)
\(\therefore \frac{\sin \theta \tan \theta(1+\cot \theta)}{\sin \theta+\cos \theta}=\frac{\tan \theta \tan \theta(1+\cot \theta)}{\tan \theta+1}\)
\(\frac{1}{\sqrt{3}}\) 

\(\frac{\tan 26^{\circ}}{\cot 64^{\circ}}=\frac{\tan \left(90^{\circ}-64^{\circ}\right)}{\cot 64^{\circ}}\)
\(=\frac{\cot 64^{\circ}}{\cot 64^{\circ}}=1\) 

tan(A + B)=\(\frac{tanA+tanB}{1-tanAtanB}\)
⇒ tan(A + B)=\(\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{2}\times\frac{1}{3}}\)
= tan(A + B)=\(\frac{\frac{3+2}{6}}{\frac{6-1}{6}}=\frac{5}{5}=1\)
tan(A + B) = 1 = tan45⁰
A + B = 45⁰
Hence proved

Here \(\sqrt { 3 } \sin { \theta } -\cos { \theta } =0\) and \({ 0 }^{ ° }<\theta <{ 90 }^{ ° }\)
\(\Rightarrow \quad \sqrt { 3 } \sin { \theta } =\cos { \theta } \)
\(\Rightarrow \quad \frac { \sin { \theta } }{ \cos { \theta } } =\frac { 1 }{ \sqrt { 3 } } \)
\(\Rightarrow \quad \tan { \theta } =\frac { 1 }{ \sqrt { 3 } } \)
\(=\tan { { 30 }^{ ° } } \left[ \because \quad \tan { \theta } =\frac { \sin { \theta } }{ \cos { \theta } } \right] \)
\(\therefore \quad \theta =\tan { { 30 }^{ ° } } \)

Given, \(\sin { A } +\sin ^{ 2 }{ A } =1\)
\(\Rightarrow \sin { A } =1-\sin ^{ 2 }{ A } \left[ \because \sin ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta } =1 \right] \)
On squaring both sides, we get
\(\sin ^{ 2 }{ A } =\cos ^{ 4 }{ A } \)
\(\Rightarrow \quad 1-\cos ^{ 2 }{ A } =\cos ^{ 4 }{ A } \Rightarrow \cos ^{ 2 }{ A } +\cos ^{ 4 }{ A } =1\)