Given (sec A + tan A)(sec B + tan B)(sec C + tan C) = (sec A - tan A)(sec B - tan B)(sec C - tan C)
Multiply both the sides by
(sec A - tan A)(sec B - tan B)(sec C - tan C)
\(\Rightarrow\)  (sec A + tan A)(sec B + tan B)(sec C + tan C) x (sec A - tan A)(sec B - tan B)(sec C - tan C)
= (sec A - tan A)²(sec B - tan B)²(sec C - tan C)²
\(\Rightarrow\) (sec²A - tan²A)(sec²B - tan²B)(sec²C - tan²C)
= (sec A - tan A)²(sec B - tan B)²(sec C - tan C)²
\(\Rightarrow\) 1 = [(sec A - tan A)(sec B - tan B)(sec C - tan C)]²
\(\Rightarrow\)  (sec A - tan A)(sec B - tan B)(sec C - tan C) = \(\pm \) 1.
Simiarly, multiply both sides by
(sec A + tan A)(sec B + tan B)(sec C + tan C),
\(\therefore\)  (sec A + tan A)(sec B + tan B)(sec C + tan C) = \(\pm \) 1.

LHS = \({ (\sin { A } +cosecA) }^{ 2 }+{ (\cos { A } +\sec { A } ) }^{ 2 }\)
\(=\sin ^{ 2 }{ A } +{ cosec }^{ 2 }A+2\sin { A } cosecA+\cos ^{ 2 }{ A } +\sec ^{ 2 }{ A } +2\cos { A } \sec { A } \)
\(\left[ \because { (a+b) }^{ 2 }={ a }^{ 2 }+{ b }^{ 2 }+2ab \right] \)
\(=(\sin ^{ 2 }{ A } +\cos ^{ 2 }{ A } )+(1+\cot ^{ 2 }{ A } )+2\sin { A } \frac { 1 }{ \sin { A } } +(1+\tan ^{ 2 }{ A } )+2\cos { A } \frac { 1 }{ \cos { A } } \)\(\left[ \because { cosec }^{ 2 }A=1+\cot ^{ 2 }{ A } ,\sec ^{ 2 }{ A } =1+\tan ^{ 2 }{ A } ,cosecA=\frac { 1 }{ \sin { A } } and\sec { A } =\frac { 1 }{ \cos { A } } \right] \)\(=1+1+\cot ^{ 2 }{ A } +2+1+\tan ^{ 2 }{ A } +2\quad \left[ \because \sin ^{ 2 }{ A } +\cos ^{ 2 }{ A } =1 \right] \)
\(=7+\tan ^{ 2 }{ A } +\cot ^{ 2 }{ A } \)
= RHS
Hence proved.

As, \(\sqrt { 3 } \sin { \theta } =\cos { \theta } \Rightarrow \tan { \theta } =\frac { 1 }{ \sqrt { 3 } } \)
\(\therefore \frac { \sin { \theta } \tan { \theta } (1+\cot { \theta } ) }{ \sin { \theta } +\cos { \theta } } =\frac { \tan { \theta } \tan { \theta } (1+\cot { \theta } ) }{ \tan { \theta } +1 } \)
\(\frac { 1+\sqrt { 3 } }{ 4 } \)

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.

(i) \({x\over y}=tan\ 45^0=1\)
⇒    x=y
\((ii)\ \ \ {4000\over y}=tan60^0=\sqrt3\)
\(⇒ y={4000\sqrt3\over 3}\)
= 2306.67m
Vertical distance between two = 4000- Y
= 1693.33m

Flag staff AB = 5 m
Let h of the height of tower BC
P is a point on the ground, PC = x
In \(\Delta PBC\), tan 30°=\(\frac{h}{x}\)
\(\frac { 1 }{ \sqrt { 3 } } =\frac { h }{ x } \)
x=\(\sqrt{3}\)h...(i)
In \(\Delta PAC\), tan60°=\(\frac{AC}{PC}\)
\(\sqrt { 3 } =\frac { 5+h }{ x } \) ...(ii)
\(\sqrt { 3 } =\frac { 5+h }{ \sqrt { 3 } h } \)
3h-h = 5
2h = 5
\(h=\frac { 5 }{ 2 } =2.5m\)

\(\cos { \theta } +\sin { \theta } =\sqrt { 2 } \cos { \theta } \)
\(\Rightarrow \quad \sin { \theta } =\cos { \theta } \left( \sqrt { 2 } -1 \right) \)
\(\Rightarrow \quad \sin { \theta } =\frac { \cos { \theta } \left( \sqrt { 2 } -1 \right) \left( \sqrt { 2 } +1 \right) }{ \left( \sqrt { 2 } +1 \right) } \)
\(\Rightarrow \quad \sin { \theta } =\frac { \cos { \theta } \left( 2-1 \right) }{ \sqrt { 2 } +1 } \)
\(\Rightarrow \quad \left( \sqrt { 2 } +1 \right) \sin { \theta } =\cos { \theta } \)
\(\Rightarrow \quad \sqrt { 2 } \sin { \theta } +\sin { \theta } =\cos { \theta } \)
\(\Rightarrow \quad \cos { \theta } -\sin { \theta } =\sqrt { 2 } \sin { \theta } \)

Given, a \(\triangle\) ABC, in which B=90⁰
In a \(\triangle\) ABC, \(tanA=\frac { Perpendicular }{ base } \)
\(=\frac { BC }{ AB } =1\)
BC=AB
Let AB=BC=k, where k is a positive number.
Now,  \(AC=\sqrt { { AB }^{ 2 }+{ BC }^{ 2 } } \)       [by unsing Pythagoras theorem]
\(=\sqrt { { (k) }^{ 2 }+{ (k) }^{ 2 } } =k\sqrt { 2 } \)
\(\therefore \quad sinA=\frac { Perpendicular }{ Hypotenuse } =\frac { BC }{ AC } =\frac { k }{ \sqrt { 2 } k } =\frac { 1 }{ \sqrt { 2 } } \)
\(cosA=\frac { Base }{ Hypotenuse } =\frac { AB }{ AC } =\frac { k }{ \sqrt { 2 } k } =\frac { 1 }{ \sqrt { 2 } } \)
Now, LHS=2sin A cos A=\(2\left( \frac { 1 }{ \sqrt { 2 } } \right) \left( \frac { 1 }{ \sqrt { 2 } } \right) =1=RHS\)
Hence proved.

\(\frac { \tan { { 29 }^{ 0 } } }{ \cot { { 61 }^{ 0 } } } =\frac { \tan { { (90 }^{ 0 }-{ 61 }^{ 0 }) } }{ \cot { { 61 }^{ 0 } } } =\frac { \cot { { 61 }^{ 0 } } }{ \cot { { 61 }^{ 0 } } } =1\)  \([\because \tan { { (90 }^{ 0 }-\theta ) } =\cot { \theta } ]\)

Let AB the tower whose height is h m.=30 m.

From ΔABC, \({AB\over BC}=tan60^0\)
 \({h\over 30}=\sqrt3\)
=30√3 m
Height of tower = 30√3 m.

Let the length of kite string AC = l m.
ㄥACB = 60°, height of kite AB = 100 m.
From ΔABC, \({AB\over BC}={\sqrt3\over 2}\)
\({100\over l}={\sqrt{3}\over 2}\)
\(l={2\times100\over\sqrt3}\)
\(={200\over\sqrt{3}}m\)
\(={200\over \sqrt3}\times{\sqrt3\over \sqrt3}\)
\(={200\sqrt{3}\over 3}m\)  

Let the height of tower be h.
From ΔABD \({h\over a}=tan\ 30^0\)

\(h=a\times{1\over \sqrt{3}}={a\over \sqrt3}\)
From ΔACD, \({h\over b}=tan\ 60^0\)
\(h=b\times\sqrt3=b\sqrt3\)
From(i) \(a=\sqrt3h\)
From (ii)\(b={h\over \sqrt3}\)
\(a\times b=\sqrt3h\times{h\over \sqrt3}\)
h²=ab
\(h=\sqrt{ab}\)

Let height of the tower AB = h metre
 AE = h-1.7
BC = DE = 20√3 m. given
In ΔADE, tan 30° =\(h-1.7\over 20\sqrt3\)
\(⇒\ {1\over \sqrt3}={h-1.7\over 20\sqrt3}\)
h-1.7 = 20
h = 20+ 1.7
= 21.7 m

\(LHS=-1+\frac { \sin { A } \sin { \left( { 90 }^{ ° }-A \right) } }{ \cot { \left( { 90 }^{ ° }-A \right) } } \)
\(\left[ \because \quad \sin { \left( { 90 }^{ ° }-\theta \right) =\cos { \theta } } \right] \)
\(\left[ \because \quad \cot { \left( { 90 }^{ ° }-\theta \right) } =\tan { \theta } \right] \)
\(=-1+\frac { \sin { A } \cos { A } }{ \tan { A } } \)
= - 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

cot 10° cot 30°  cot 80° = cot(90° - 80°) cot 30° cot 80°
= tan 80° cot 30° \(\frac { 1 }{ \tan { { 80 }^{ ° } } } \)
= cot 30°
\(=\sqrt{3}\)

sin (90° - 2A) = sin (A - 15°)
\(\Rightarrow\)  90° - 2A = A - 15
\(\Rightarrow\)  3A = 105°
\(\therefore\)  A = 35°

Given tan 2A = cot (A + 60°)
\(\Rightarrow\)  cot(90 - 2A) = cot (A + 60)
\(\Rightarrow\)  90 - 2A = A + 60°
\(\Rightarrow\)  3A = 30°
\(\therefore\)  A = 10°

A + B + C = 180°
B + C = 180° - A
\(\therefore \quad \cos { \left( \frac { B+C }{ 2 } \right) } =\cos { \left[ \frac { { 180 }^{ ° }-A }{ 2 } \right] } \)
\(=\cos { \left( { 90 }^{ ° }-\frac { A }{ 2 } \right) } \)
\(=\sin { \frac { A }{ 2 } } \)

Given, \(\angle CBA=90^0,\angle BRS=90^0,\angle BSC=90^0,\) BC = 7.5 cm, RS = 5 cm, BR = 6 cm and AB = 18 cm
Then, AR = AB - RB = 18 - 6 = 12 cm

In \(\triangle ABC,\)  by using Pythagoras theorem, we get  
AS² = AR² + RS² = (12)² + (5)²
AS² = 144 + 25 = 169
\(\Rightarrow\) AS = \(\sqrt 169\)
\(\therefore\)  AS = 13 cm [since, side cannot be negative]
\((i)\quad In\quad \triangle BRS,\ tan{ x }^{ 0 }=\frac { P }{ B } =\frac { BR }{ RS } =\frac { 6 }{ 5 } \quad \)
\((ii)\quad In\quad \triangle ARS,\quad sin{ y }^{ 0 }=\frac { P }{ H } =\frac { SR }{ AS } =\frac { 5 }{ 13 } \)
\((iii)\quad In\quad \triangle ARS,\quad cos{ y }^{ 0 }=\frac { AR }{ AS } =\frac { 12 }{ 13 } \)

\(\tan { \theta } =\frac { 1 }{ \sqrt { 3 } } \Rightarrow \theta ={ 30 }^{ 0 }\)