LHS = \(cot{ \left( 7\frac { 1 }{ 2 } \right) }^{ ° }\)
= \(\frac { cos{ 7\frac { 1 }{ 2 } }^{ ° } }{ sin{ 7\frac { 1 }{ 2 } }^{ ° } } \)
Multiplying the numerator and denominator by 2sin(\({ 7\frac { 1 }{ 2 } }^{ ° }\))
\(\frac { 2sin{ 7\frac { 1 }{ 2 } }^{ ° }cos{ 7\frac { 1 }{ 2 } }^{ ° } }{ 2{ sin }^{ 2 }{ 7\frac { 1 }{ 2 } }^{ ° } } =\frac { sin15° }{ 1-cos15° } \)
\(\frac { sin\left( 45-30° \right) }{ 1-cos\left( 45-30° \right) } =\frac { sin45cos30-cos45°sin30° }{ 1-\left( cos45°cos30°+sin45sin30 \right) } \)
= \(\frac { \frac { 1 }{ \sqrt { 2 } } .\frac { \sqrt { 3 } }{ 2 } -\frac { 1 }{ \sqrt { 2 } } .\frac { 1 }{ 2 } }{ 1-\left( \frac { 1 }{ 2 } .\frac { \sqrt { 3 } }{ 2 } +\frac { 1 }{ \sqrt { 2 } } .\frac { 1 }{ 2 } \right) } =\frac { \sqrt { 3 } -1 }{ 2\sqrt { 2 } } /1-\left( \frac { \sqrt { 3 } +1 }{ 2\sqrt { 2 } } \right) \)
= \(\frac { \sqrt { 3 } -1 }{ 2\sqrt { 2 } } \times \frac { 2\sqrt { 2 } }{ 2\sqrt { 2 } -\sqrt { 3 } -1 } =\frac { \sqrt { 3 } -1 }{ 2\sqrt { 2 } -\sqrt { 3 } -1 } \)
= \(\frac { \sqrt { 3 } -1 }{ 2\sqrt { 2 } -\sqrt { 3 } -1 } \times \frac { 2\sqrt { 2 } +\sqrt { 3 } +1 }{ 2\sqrt { 2 } +\sqrt { 3 } +1 } \)
\(\frac { \left( \sqrt { 3 } -1 \right) \left( 2\sqrt { 2 } +\sqrt { 3 } +1 \right) }{ { \left( 2\sqrt { 2 } \right) }^{ 2 }-{ \left( \sqrt { 3 } +1 \right) }^{ 2 } } =\frac { 2\sqrt { 6 } +3-2\sqrt { 3 } -\sqrt { 3 } -1 }{ 8-\left( 1+3+2\sqrt { 3 } \right) } \)
= \(\frac { 2\sqrt { 6 } +2-2\sqrt { 2 } }{ 4-2\sqrt { 3 } } =\frac { \sqrt { 6 } +1-\sqrt { 2 } }{ 2-\sqrt { 3 } } \times \frac { 2+\sqrt { 3 } }{ 2+\sqrt { 3 } } \)
= \(\frac { 2\sqrt { 6 } +2-2\sqrt { 2 } +\sqrt { 18 } +\sqrt { 3 } -\sqrt { 6 } }{ 4-2\sqrt { 3 } } =\frac { \sqrt { 6 } +1-\sqrt { 2 } }{ 2-\sqrt { 3 } } \times \frac { 2+\sqrt { 3 } }{ 2+\sqrt { 3 } } \)
= \(\frac { 2\sqrt { 6 } +2-2\sqrt { 2 } +\sqrt { 18 } +\sqrt { 3 } -\sqrt { 6 } }{ 4-3 } =2\sqrt { 6 } +2-2\sqrt { 2 } +3\sqrt { 2 } +\sqrt { 3 } -\sqrt { 6 } \)
= \(\sqrt { 6 } +\sqrt { 3 } +2-2\sqrt { 2 } +3\sqrt { 2 } =\sqrt { 6 } +\sqrt { 3 } +2+2\sqrt { 2 } \)
= \(\sqrt { 2 } +\sqrt { 3 } +\sqrt { 4 } +\sqrt { 6 } \) = RHS
Hence proved.

LHS = cos 2A + cos 2B + cos 2C
= 2cos \(\left( \frac { 2A+2B }{ 2 } \right) \)cos\(\left( \frac { 2A-2B }{ 2 } \right) \)1-2 sin²C .
= 2cos(A + B) cos(A - B) +1-2sin²C 
= 2cos\(\left( \frac { \pi }{ 2 } -C \right) \)cos(A-B) + 2sin²C + 1
= 1 + 2 sin C cos(A - B)-2sin²C
= 1 + 2 sinC\(\left[ cos(A-B)+sin\left( \frac { \pi }{ 2 } (A+B) \right) \right] \)
= 1 + 2sinC [cos(A - B)-cos(A + B)]
= 1 + 2 sin C(2sinA sin B)
= 1 + 4 Sin A sin B sin C

sin (A + B) sin ( A - B ) = sin² A - sin² B.
LHS = sin (A + B) sin (A - B)
= [sin A cos B + cos A sin B] [ sin A cos B - cos A sin B]
= sin² A cos ² B - (1 - sin² A) sin² B
= sin² (1 - sin² B) - (1 - sin² A) sin² B
= sin² A - sin² A sin² B - sin² B + sin² A sin² B
= sin² A - sin² B = RHS
Hence proved.

Given \(\frac { cos^{ 4 }\alpha }{ { cos }^{ 2 }\beta } +\frac { { sin }^{ 4 }\alpha }{ { sin }^{ 2 }\beta } \)
⇒ cos⁴\(\alpha\)sin²β + sin⁴\(\alpha\)cos²β = cos²βsin²β
⇒ cos⁴\(\alpha\)(1-cos²β) + cos²β(1-cos²\(\alpha\))2 = cos²β(1-cos²β)

⇒ cos⁴\(\alpha\)-2cos²\(\alpha\)cos²β+cos⁴β = 0
⇒ (cos²\(\alpha\)-cos²β)² = 0
⇒ cos²\(\alpha\)-cos²β = 0
⇒ cos²\(\alpha\) = cos²β...(1)
⇒ 1-sin²\(\alpha\) = 1-sin²β
⇒ sin²\(\alpha\) = sin^-2β
\(\frac { cos^{ 4 }\alpha }{ { cos }^{ 2 }\beta } +\frac { { sin }^{ 4 }\alpha }{ { sin }^{ 2 }\beta } =1\)
LHS = \(\frac { { cos }^{ 4 }\beta }{ { cos }^{ 2 }\alpha } +\frac { { sin }^{ 4 }\beta }{ { sin }^{ 2 }\alpha } \)
= \(\frac { { cos }^{ 2 }\beta { cos }^{ 2 }\beta }{ { cos }^{ 2 }\alpha } +\frac { { sin }^{ 2 }\beta { sin }^{ 2 }\beta }{ { sin }^{ 2 }\alpha } \)
\(={ cos }^{ 2 }\beta +{ sin }^{ 2 }\beta =1\) = RHS

Let s be the position of the satellite, E be the position of the earth station and C be the centre of the earth.

Given CE = r, CS= Rand SE = d
Given LSCE = \(\alpha\)
In ΔSCE, applying cosine rule, we get
d² = -2 + R² - 2 (r)(R)cos\(\alpha\)
d² = r²+ R² - 2r.R.cos \(\alpha\)
Dividing by R² throughout we get,
\(\frac { { d }^{ 2 } }{ { R }^{ 2 } } =\frac { { r }^{ 2 } }{ { R }^{ 2 } } +\frac { { R }^{ 2 } }{ { R }^{ 2 } } -\frac { 2rR }{ { R }^{ 2 } } cos\alpha \)
\(\Rightarrow \frac { { d }^{ 2 } }{ { R }^{ 2 } } =\frac { { r }^{ 2 } }{ { R }^{ 2 } } +1-\frac { 2r }{ R } cos\alpha \)
\(\Rightarrow { d }^{ 2 }={ R }^{ 2 }\left[ 1+\frac { { r }^{ 2 } }{ { R }^{ 2 } } -\frac { 2r }{ R } cos\alpha \right] \)
Taking positive square root both sides we get,
\(d=R\sqrt { 1+\frac { { r }^{ 2 } }{ { R }^{ 2 } } -\frac { 2r }{ R } } cos\alpha \)
d = \(R​​\sqrt { 1+\left( \frac { r }{ R } \right) ^{ 2 }-2\frac { r }{ R } cos\alpha } \)
Hence proved.

Let C be the position of the target and A and B be the positions of the fighter jet

Given ㄥBAC = 30, ㄥABC = 45
ஃ ㄥC = 180 - (30 - 45) = 180 - 75 = 105
Given AB = 100 km
Using sine formula,
\(\frac { a }{ sinA } =\frac { c }{ sinC } \)
\(\Rightarrow \frac { a }{ sin30° } =\frac { 100 }{ sin105° } \)
\(\Rightarrow \frac { a }{ \frac { 1 }{ 2 } } =\frac { 100 }{ sin105° } \Rightarrow 2a=\frac { 100 }{ sin105° } \Rightarrow a=\frac { 50 }{ sin105° } \)
Now, sin 105° = sin (60 + 45) = sin 60 cos 45 + cos 60 sin 45
= \(\frac { \sqrt { 3 } }{ 2 } .\frac { 1 }{ \sqrt { 2 } } +\frac { 1 }{ 2 } .\frac { 1 }{ \sqrt { 2 } } =\frac { \sqrt { 3 } +1 }{ 2\sqrt { 2 } } \)
Substituting (2) in (1) we get,
a = \(\frac { 50 }{ \frac { \sqrt { 3 } +1 }{ 2\sqrt { 2 } } } \Rightarrow a=\frac { 50\left( 2\sqrt { 2 } \right) }{ \sqrt { 3 } +1 } =\frac { 100\sqrt { 2 } }{ \sqrt { 3 } +1 } \times \frac { \sqrt { 3 } -1 }{ \sqrt { 3 } -1 } \)
a = \(\frac { 100\left( \sqrt { 6 } -\sqrt { 2 } \right) }{ 3-1 } =50\left( \sqrt { 6 } -\sqrt { 2 } \right) km\)

cot(660°) = cot (360 \(\times\) 2 - 60)
\(=-cot 60°=-\frac{1}{\sqrt 3}\)

Let sin^-1\(({\sqrt{3}\over2})\) = y, where -\({\pi\over 2}\le y \le {\pi\over 2}\)
\(\Rightarrow \) sin y = \({\sqrt{3}\over2}\) = sin \({\pi\over3}\Rightarrow y ={\pi\over3}\)
Thus, the principal value of sin^-I \(({\sqrt{3}\over2})\) = \({\pi\over 3}\)

Given trigonometric equation is
 \(\sqrt { 3 } \) sec 2x = 2
⇒ sec 2x = \(\frac { 2 }{ \sqrt { 3 } } \)
⇒ cos 2x = \(\frac { \sqrt { 3 } }{ 2 } \)
⇒ cos 2x = cos\(\left( \frac { \pi }{ 6 } \right) \)    \(\left[ \because sec2x=\frac { 1 }{ cos2x } \right] \)
⇒ \(2x=2n\pi \pm \left( \frac { \pi }{ 6 } \right) ,n\in Z\)    \(\left[ \because cos\frac { \pi }{ 6 } =\frac { \sqrt { 3 } }{ 2 } \right] \)
⇒ x = \(n\pi \pm \frac { \pi }{ 12 } ,n\in Z\)

\(\frac { cos(2\pi +x)cosec(2\pi +x)tan\left( \frac { \pi }{ 2 } +x \right) }{ sec\left( \frac { \pi }{ 2 } +x \right) cos.cot(\pi +x) } \) = 1
LHS=\(\frac { cos(2\pi +x)cosec(2\pi +x)tan\left( \frac { \pi }{ 2 } +x \right) }{ sec\left( \frac { \pi }{ 2 } +x \right) cos.cot(\pi +x) } \)
cos(2\(\pi \) + x) = cos\(\pi \)
cosec(2\(\pi \) + x)cosec x
\(tan\left( \frac { \pi }{ 2 } +x \right) \) = -cot x
\(sec\left( \frac { \pi }{ 2 } +x \right) \)= -cosec x
cot(\(\pi \)+x) = +cot x
LHS = \(\frac { cosx.cosecx.(-cotx) }{ (-cosecx).cosx(+cotx) } \) = 1

\(\frac { \pi }{ 3 } \)
\(\frac { \pi }{ 3 } \) = \(\frac { \pi }{ 3 } \) \(\times\)\(\frac { 180 }{ \pi } \) = 60^o

30⁰
30⁰ = 30 \(\times\) \(\frac { \pi }{ 180 } =\frac { \pi }{ 6 } \)

\(\sin \alpha+\cos \alpha=b \Rightarrow(\sin \alpha+\cos \alpha)^{2} =b^{2} \)
\(\sin ^{2} \alpha+\cos ^{2} \alpha+2 \sin \alpha \cos \alpha =b^{2} \)
\(\sin 2 \alpha =b^{2}-1 \)
Since \(-1<\sin 2 \alpha \leq 1 \)
\(-1 \leq \mathrm{b}^{2}-1 \leq 1 \)
\(\mathrm{~b}^{2}-1 \leq 1 \)
\(\mathrm{~b}^{2} \leq 2 \)
\(\text { This is possible if } \mathrm{b} \leq \sqrt{2}\)
\(\sin 2 \alpha=b^{2}-1, \text { if } b \leq \sqrt{2}\)

\(\mathrm{f}_{4}(x)-\mathrm{f}_{6}(x)=\frac{1}{4}\left[\sin ^{4} x+\cos ^{4} x\right]-\frac{1}{6}\left[\sin ^{6} x+\cos ^{6} x\right] \)
\(=\frac{1}{4}\left[\left(\sin ^{2} x+\cos ^{2} x\right)^{2}-2 \sin ^{2} x \cos ^{2} x\right]-\frac{1}{6}\left[\left(\sin ^{2} x+\cos ^{2} x\right)^{3}\right. \)
\(\left.=-3 \sin ^{2} x \cos ^{2} x+\left(\sin ^{2} x+\cos ^{2} x\right)\right] \)
\(=\frac{1}{4}\left(1-2 \sin ^{2} x \cos ^{2} x\right)-\frac{1}{6}\left(1-3 \sin ^{2} x \cos ^{2} x\right) \)
\(=\frac{1}{4}-\frac{1}{2} \sin ^{2} x \cos ^{2} x-\frac{1}{6}+\frac{1}{2} \sin ^{2} x \cos ^{2} x \)
\(=\frac{1}{4}-\frac{1}{6} \)
\(=\frac{3-2}{12}=\frac{1}{12} \)

\(x =\frac{1}{\cos 80^{\circ}}-\frac{\sqrt{3}}{\sin 80^{\circ}} \)
\(=\frac{\sin 80^{\circ}-\sqrt{3} \cos 80^{\circ}}{\sin 80^{\circ} \cos 80^{\circ}} \)
\(\frac{x}{2} =\frac{\frac{1}{2} \sin 80^{\circ}-\frac{\sqrt{3}}{2} \cos 80^{\circ}}{\frac{1}{2} \sin 80^{\circ} \cos 80^{\circ}} \)
\(=\frac{\sin 80^{\circ} \cos 60^{\circ}-\cos 80^{\circ} \sin 60^{\circ}}{\frac{1}{2} \sin 80^{\circ} \cos 80^{\circ}} \)
\(=\frac{\sin 20^{\circ}}{\frac{1}{2} \sin 80^{\circ} \cos 80^{\circ}} \)
\(=\frac{\sin 160^{\circ}}{\frac{1}{2} \sin 80^{\circ} \cos 80^{\circ}} \)
\(=\frac{2 \sin 80^{\circ} \cos 80^{\circ}}{\frac{1}{2} \sin 80^{\circ} \cos 80^{\circ}}=4 \)

LHS = b cos A + c cos A + c cos B + a cos B + a cos C + b cos C
= b cos C + c cos B + c cos A + a cos C + b cos A + a cos B
= a + b + c [by projection formula]

We have sin 34° + cos 64° - cos 4°\(=sin34^0-2sin(\frac{64^0+4^0}{2})sin(\frac{64^0-4^0}{2})\)
= sin 34° - 2 sin 34° sin 30° = 0

A circular metallic plate is in the form of a cylinder with radius 8 cm, and height = 6 mm = \(\frac { 6 }{ 10 } \) Given that volume of cylinder = Volume of sector 
\(\pi { r }^{ 2 }h=\frac { \theta }{ 360 } \times \pi \times { r }^{ 2 }\times h\) 
\( \Rightarrow \pi\left(80^{2}\right) \times 6 =\frac{\theta}{360^{\circ}} \times(160)^{2} \times 4 \)
\(\theta =\frac{3 \times 180^{\circ}}{8} \)
\(\theta =\frac{3 \pi}{8} \text { radians. }\)