sin 67⁰ + cos 75⁰ = \(\sin { ({ 90 }^{ 0 }-{ 23 }^{ 0 }) } +\cos { ({ 90 }^{ 0 }-{ 15 }^{ 0 }) } \)
\(=\cos { { 23 }^{ 0 } } +\sin { { 15 }^{ 0 } } \left[ \because \sin { ({ 90 }^{ 0 }-\theta ) } =\cos { \theta } ,\cos { ({ 90 }^{ 0 }-\theta ) } =\sin { \theta } \right] \) 

LHS = \(\sqrt { \frac { 1+\sin { A } }{ 1-\sin { A } } \times \frac { 1+\sin { A } }{ 1+\sin { A } } } +\sqrt { \frac { 1-\sin { A } }{ 1+\sin { A } } \times \frac { 1-\sin { A } }{ 1-\sin { A } } } \)
\(=\frac { 1+\sin { A } }{ \cos { A } } +\frac { 1-\sin { A } }{ \cos { A } } =2\sec { A } \) 

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

\(\sin { B } =\frac { 12 }{ 13 } ,\cos { C } =\frac { 12 }{ 13 } ,\quad \tan { B } =\frac { 12 }{ 5 } \)

We have, \(sinAcosA-\frac { sinAcos(90°-A)cosA }{ sec(90°-A) } -\frac { cosAsin(90°-A)sinA }{ cosec(90°-A) } \)
\(=sinAcosA-\frac { sinAsinAcosA }{ cosecA } -\frac { cosAcosAsinA }{ secA } \)
\(=sinAcosA-sinAcosA-sinAcosA\left( \frac { sinA }{ cosecA } +\frac { cosA }{ secA } \right) \)
\(=sinAcosA-sinAcosA({ sin }^{ 2 }A+{ cos }^{ 2 }A)\)
\(=sinAcosA-sinAcosA\)
=0

cos²  \(\theta\)= 1 - sin ²\(\theta\) to find cos A and cos B. 
\( \cos { A } =\frac { 2 }{ \sqrt { 5 } } ,\cos { B } =\frac { 3 }{ \sqrt { 10 } } \)  

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

\(\cos { { 53 }^{ 0 } } -\sin { { 37 }^{ 0 } } \)
\(=\cos { { (90 }^{ 0 }-{ 37 }^{ 0 }) } -\sin { { 37 }^{ 0 } } \)
\(=\sin { { 37 }^{ 0 } } -\sin { { 37 }^{ 0 } } =0\quad \left[ \because \cos { { (90 }^{ 0 }-\theta ) } =\sin { \theta } \right] \) 

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

Let C be the point where the ball is
ㄥC = 60° (alternate angles)
In ΔABC, \(tan\ 60^0={AB\over BC}\)
⇒ \(\sqrt3={20\over x}\)
⇒ \(x={20\over \sqrt3}\)
\(=20\left(\sqrt3\over3 \right)\)
Hence, required distance is
1 = 11.53m

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.

Given AB = 4m, AC = 8 m.
Angle of elevation = ㄥACB =
From ΔABC, \({AB\over AC}=sin\theta\)
\(⇒ {4\over 8}=sin\theta\)
\(⇒ sin\theta={1\over 2}=in30^0\)
\(⇒ \theta=30^0\)

From the triangle ABC
\({h\over 28.2}=tan60^0\)
h = 28.2 x tan 60°
= 28.2 x \(\sqrt3\)
= 28.2\(\sqrt3\) + 1.2m
Height ofthe tower = AC+CD
= 50 m

\(2\cot ^{ 2 }{ A } -1=2\left( { cosec }^{ 2 }A-1 \right) -1\)
\(=\frac { 2 }{ \sin ^{ 2 }{ A } } -3\)
\(\left( \because \quad \cot ^{ 2 }{ \theta } =-1+{ cosec }^{ 2 }\theta \right) \)
\(=\frac { 2 }{ { \left( \frac { \sqrt { 3 } }{ 2 } \right) }^{ 2 } } -3\)
\(2\cot ^{ 2 }{ A } -1=\frac { 8 }{ 3 } -3=\frac { -1 }{ 3 } \)

= sin² 41° + sin² 49°.
= sin² (90° - 49°) + sin² 49°
= cos² 49° + sin² 49°
= 1      \(\left[ \because \quad \cos ^{ 2 }{ \theta } +\sin ^{ 2 }{ \theta } =1 \right] \)

\(\frac { \cos { { 45 }^{ ° } } }{ \sec { { 30 }^{ ° } } } +\frac { 1 }{ \sec { { 60 }^{ ° } } } =\frac { \frac { 1 }{ \sqrt { 2 } } }{ \frac { 2 }{ \sqrt { 3 } } } +\frac { 1 }{ 2 } \)
\(=\frac { 1 }{ \sqrt { 2 } } \times \frac { \sqrt { 3 } }{ 2 } +\frac { 1 }{ 2 } \)
\(=\frac { \sqrt { 6 } }{ 4 } +\frac { 1 }{ 2 } \)
\(=\frac { \sqrt { 6 } +2 }{ 4 } \)

= tan² 10° - cot² 80°
= tan²(90° - 80°) - cot² 80°
\(\left[ \therefore \quad \tan { \left( { 90 }^{ ° }-\theta \right) =\cot { \theta } } \right] \)
= cot² 80° - cot² 80°
= 0

So, the given equation is rewritten as
\(\sqrt{3} x^{2}-4 x+\sqrt{3}=0\)
Using quadratic formula, we get
\(x=\sqrt{3}, \frac{1}{\sqrt{3}} \text { i.e. } \cot \theta=\sqrt{3}, \frac{1}{\sqrt{3}}\) \(\text { of, } \tan \theta=\frac{1}{\sqrt{3}}, \sqrt{3} \text { . }\)
\(\frac{10}{3}\) 

\(\cos { \theta } =\frac { x-k }{ a } \quad and\quad \sin { \theta } =\frac { h+b }{ y } \)