LHS = (1 + sec2θ)(1 + sec4θ)...(1 + sec2ⁿθ)
= \(\left( 1+\frac { 1 }{ cos2\theta } \right) \left( 1+\frac { 1 }{ cos4\theta } \right) ...\left( 1+\frac { 1 }{ cos{ 2 }^{ n }\theta } \right) \)
= \(\frac { \left( cos2\theta +1 \right) \left( cos4\theta +1 \right) ...\left( cos{ 2 }^{ n }\theta +1 \right) }{ cos2\theta .cos4\theta ...cos{ 2 }^{ n }\theta } \)
= \(\frac { \left( 2{ cos }^{ 2 }\theta \right) \left( 2{ cos }^{ 2 }2\theta \right) ...\left( 2{ cos }^{ 2 }{ 2 }^{ n-1 }x \right) }{ cos2\theta \quad cos{ 2 }^{ 2 }\theta \quad cos{ 2 }^{ n }\theta } \)
= \(\frac { { 2 }^{ n }cos\theta }{ { cos }^{ 2n }\theta } \left[ cosx.cos2x.cos{ 2 }^{ 2 }x...cos{ 2 }^{ n-1 }x \right] \)
= \(\frac { { 2 }^{ n }cos\theta }{ { cos }^{ 2n }\theta } \times \frac { cos{ 2 }^{ n }\theta }{ { 2 }^{ n }sinx } \)
= tan²ⁿθ.cotθ
= RHS

Given cos(\(\alpha \) - β)+cos(β - \(\gamma \))+cos(\(\gamma \) - \(\alpha \)) = - \(\frac{3}{2}\)
\(cos\alpha cos\beta +sin\alpha sin\beta +cos\beta cos\gamma +sin\beta sin\gamma +cos\gamma cos\alpha +sin\alpha sin\gamma =-\frac { 3 }{ 2 } \)
\(2\left[ cos\alpha cos\beta +cos\beta cos\gamma +cos\gamma cos\alpha +sin\alpha sin\beta +sin\beta sin\gamma +sin\alpha sin\gamma \right] =-3\)
\(\left( 2cos\alpha cos\beta +2cos\beta cos\gamma +2cos\gamma cos\alpha \right) +\left( 2sin\alpha sin\beta +2sin\beta sin\gamma +2sin\alpha sin\gamma \right) +3=0\)
\(\left( 2cos\alpha cos\beta +2cos\beta cos\gamma +2cos\gamma cos\alpha \right) +\left( 2sin\alpha sin\beta +2sin\beta sin\gamma +2sin\alpha sin\gamma \right) +\left( { cos }^{ 2 }\alpha +{ sin }^{ 2 }\alpha \right) +\left( { cos }^{ 2 }\beta +{ sin }^{ 2 }\beta \right) +\left( { cos }^{ 2 }\gamma +{ sin }^{ 2 }\gamma \right) =0\)
\({ \left( cos\alpha +cos\beta +cos\gamma \right) }^{ 2 }+{ \left( sin\alpha +sin\beta +sin\gamma \right) }^{ 2 }=0\)
\(cos\alpha +cos\beta +cos\gamma =0\) and \(sin\alpha +sin\beta +sin\gamma =0\)

Given \(x=\sum _{ n=0 }^{ \infty }{ { cos }^{ 2n }\theta } ={ cos }^{ 0 }\theta +{ cos }^{ 2 }\theta +{ cos }^{ 4 }\theta +...\)
= \(1+{ cos }^{ 2 }\theta +{ cos }^{ 4 }\theta +...\)
= \(1+{ \left( cos\theta \right) }^{ 2 }+{ \left( { cos }^{ 2 }\theta \right) }^{ 2 }+...=\frac { 1 }{ 1-{ cos }^{ 2 }\theta } \)
= \(\frac { 1 }{ { sin }^{ 2 }\theta } \) ...(1)
\(y=\sum _{ n=0 }^{ \infty }{ { sin }^{ 2n }\theta } ={ sin }^{ 0 }\theta +{ sin }^{ 2 }\theta +{ sin }^{ 4 }\theta +...\)
= \(1+{ sin }^{ 2 }\theta +{ sin }^{ 4 }\theta +...\)
= \(1+{ \left( sin\theta \right) }^{ 2 }+{ \left( { sin }^{ 2 }\theta \right) }^{ 2 }+...=\frac { 1 }{ 1-{ sin }^{ 2 }\theta } =\frac { 1 }{ { cos }^{ 2 }\theta } \)...(2)
Similarly z = \(\sum _{ n=0 }^{ \infty }{ { cos }^{ 2n }\theta } { sin }^{ 2n }\theta ={ cos }^{ 0 }\theta { sin }^{ 0 }\theta +{ cos }^{ 2 }\theta { sin }^{ 2 }\theta +{ cos }^{ 4 }\theta +{ sin }^{ 4 }\theta ...\)
= \(1+{ \left( sin\theta cos\theta \right) }^{ 2 }+{ \left( { sin }^{ 2 }\theta { cos }^{ 2 }\theta \right) }^{ 2 }+...\)
= \(\frac { 1 }{ 1-{ sin }^{ 2 }\theta { cos }^{ 2 }\theta } \) ...(3)
Now x + y + z = \(\frac { 1 }{ { sin }^{ 2 }\theta } +\frac { 1 }{ { cos }^{ 2 }\theta } +\frac { 1 }{ 1-{ sin }^{ 2 }\theta { cos }\theta } \)(using 1, 2 and 3)
= \(\frac { { cos }^{ 2 }\theta \left( 1-{ sin }^{ 2 }\theta { cos }^{ 2 }\theta \right) +{ sin }^{ 2 }\theta \left( 1-{ sin }^{ 2 }\theta { cos }^{ 2 }\theta \right) +{ sin }^{ 2 }\theta { cos }^{ 2 }\theta }{ { sin }^{ 2 }\theta { cos }^{ 2 }\theta \left( 1-{ sin }^{ 2 }\theta { cos }^{ 2 }\theta \right) } \)
= \(\frac { { cos }^{ 2 }\theta -{ sin }^{ 2 }\theta { cos }^{ 4 }\theta +{ sin }^{ 2 }\theta --{ sin }^{ 4 }\theta { cos }^{ 4 }\theta +{ sin }^{ 2 }\theta { cos }^{ 2 }\theta }{ { sin }^{ 2 }\theta { cos }^{ 2 }\theta \left( 1-{ sin }^{ 2 }\theta { cos }^{ 2 }\theta \right) } \)
= \(\frac { 1-\left( { sin }^{ 2 }\theta { cos }^{ 4 }\theta +{ sin }^{ 2 }\theta { cos }^{ 4 }\theta \right) +{ sin }^{ 2 }\theta { cos }^{ 2 }\theta }{ { sin }^{ 2 }\theta { cos }^{ 2 }\theta \left( 1-{ sin }^{ 2 }\theta { cos }^{ 2 }\theta \right) } \)
= \(\frac { 1-{ sin }^{ 2 }\theta { cos }^{ 2 }\theta (1)+{ sin }^{ 2 }\theta { cos }^{ 2 }\theta }{ { sin }^{ 2 }\theta { cos }^{ 2 }\theta \left( 1-{ sin }^{ 2 }\theta { cos }^{ 2 }\theta \right) } \)
= \(\frac { 1 }{ { sin }^{ 2 }\theta { cos }^{ 2 }\theta \left( 1-{ sin }^{ 2 }\theta { cos }^{ 2 }\theta \right) } =\frac { 1 }{ { sin }^{ 2 }\theta } \times \frac { 1 }{ { cos }^{ 2 }\theta } \times \frac { 1 }{ 1-{ sin }^{ 2 }\theta { cos }^{ 2 }\theta } \)
ஃ x + y + z = xyz
Hence proved.

Given \(\frac { cos^{ 4 }\alpha }{ { cos }^{ 2 }\beta } +\frac { { sin }^{ 4 }\alpha }{ { sin }^{ 2 }\beta } \)
⇒ cos⁴\(\alpha\)sin²β + sin⁴\(\alpha\)cos²β = cos²β sin²β
⇒ 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
sin⁴\(\alpha\)+sin⁴β = 2sin²\(\alpha\)sin²β
LHS = sin⁴\(\alpha\) + sin⁴β = (sin²-sin²β)² + 2sin²\(\alpha\)sin²β
= 2sin²\(\alpha\)sin²β = [∵ sin θ sin²\(\alpha\) = sin²β, from (2)]
= RHS
Hence proved.

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 = \(\frac { 100\left( \sqrt { 6 } -\sqrt { 2 } \right) }{ 3-1 } =50\left( \sqrt { 6 } -\sqrt { 2 } \right) km\)

Sum of the roots = sin 15^o + cos 15^o
= sin (45 - 30) + cos (45 - 30)
= (sin 45 cos 30 - cos 45 sin 30) + (cos 45 cos 30 + sin 45 sin 30)
= \(\left( \frac { 1 }{ \sqrt { 2 } } .\frac { \sqrt { 3 } }{ 2 } -\frac { 1 }{ \sqrt { 2 } } .\frac { 1 }{ 2 } \right) +\left( \frac { 1 }{ \sqrt { 2 } } .\frac { \sqrt { 3 } }{ 2 } -\frac { 1 }{ \sqrt { 2 } } .\frac { 1 }{ 2 } \right) \)
= \(\frac { \sqrt { 3 } }{ 2\sqrt { 2 } } -\frac { 1 }{ 2\sqrt { 2 } } +\frac { \sqrt { 3 } }{ 2\sqrt { 2 } } +\frac { 1 }{ 2\sqrt { 2 } } =\frac { 2\sqrt { 3 } }{ 2\sqrt { 2 } } =\frac { \sqrt { 3 } }{ \sqrt { 2 } } \times \frac { \sqrt { 2 } }{ \sqrt { 2 } } =\frac { \sqrt { 6 } }{ 2 } \)
Product of the roots = sin 15^o cos 15^o  = \(\frac { 2\sin { { 15 }^{ o } } \cos { { 15 }^{ o } } }{ 2 } \)
= \(\frac { \sin { { 30 }^{ o } } }{ 2 } =\frac { 1 }{ 2 } \times \frac { 1 }{ 2 } =\frac { 1 }{ 4 } \left[ \begin{matrix} since & \sin { 2A } \\ =2\sin { A } & \cos { A } \end{matrix} \right] \)
Required equation is
x² - x (sum of the roots) + product of the roots = 0
\(\Rightarrow { x }^{ 2 }-x\left( \frac { \sqrt { 6 } }{ 2 } \right) +\frac { 1 }{ 4 } =0\)
Multiplying by 4 we get \({ 4x }^{ 2 }-2\sqrt { 6x } +1=0\)

Since 0\(\frac { \pi }{ 2 } \) and 0\(\frac { \pi }{ 2 } \), A, B lies in the I quadrant
ஃ All the trigonometric ratios are positive

= \(\sqrt { { 5 }^{ 2 }-{ 3 }^{ 2 } } =4\)

= \(\sqrt { { 41 }^{ 2 }-{ 9 }^{ 2 } } \)
= 40
\(sinA=\frac { 3 }{ 5 } \quad sinB=\frac { 40 }{ 41 } \)
\(cosA=\frac { 4 }{ 5 } \quad cosB=\frac { 9 }{ 41 } \)
cos (A - B) = cosA cosB + sinAsinB
= \(\frac { 4 }{ 5 } .\frac { 9 }{ 41 } +\frac { 3 }{ 5 } .\frac { 40 }{ 41 } \)
= \(\frac { 36 }{ 205 } +\frac { 120 }{ 205 } =\frac { 156 }{ 205 } \)

\(3 \mathrm{a}=12 \Rightarrow \mathrm{a}=4\)
Maximum area is obtained when it is an equilateral triangle with side 4 m each

\(\cos p \theta+\cos q \theta=0 \)
\(2 \cos \left(\frac{p+q}{2}\right) \theta \cdot \cos \left(\frac{p-q}{2}\right) \theta=0 \)
\(\Rightarrow 2 \cos \left(\frac{p+q}{2}\right) \theta=0 \)
\(\cos \left(\frac{\mathrm{p}-\mathrm{q}}{2}\right) \theta=0\)
\(\text { Principal angle is } \pi / 2\)
\(\therefore\left(\frac{p+q}{2}\right) \theta =n \pi \pm \pi / 2 \)
\(\Rightarrow(p+q) \cdot \theta =2 n \pi \pm \pi \)
\(\theta =\frac{\pi(2 n \pm 1)}{p+q} \)
\(\text { Similarly, } \theta=\frac{\pi(2 n \pm 1)}{p-q}\)

\(\sqrt{2+2 \cos 4 \theta} =\sqrt{2+2\left(2 \cos ^{2} 2 \theta-1\right)} \)
\(=\sqrt{4 \cos ^{2} 2 \theta} \)
\(=2 \cos 2 \theta \)
\(\sqrt{2+\sqrt{2+2 \cos 4 \theta}} =\sqrt{2+2 \cos 2 \theta} \)
\(=\sqrt{2+2\left(2 \cos ^{2} \theta-1\right)} \)
\(=\sqrt{4 \cos ^{2} \theta} \)
\(=\pm 2 \cos \theta \)
\(\Rightarrow \pi<2 \theta<\frac{3 \pi}{2} \)
\(\Rightarrow \frac{\pi}{2}<\theta<\frac{3 \pi}{4} \text { in II quadrant. } \)
\(\therefore \sqrt{2+\sqrt{2+2 \cos 4 \theta}}=-2 \cos \theta\)

\(\cos 28^{\circ}+\sin 28^{\circ} =\mathrm{k}^{3} \)
\(\cos 28^{\circ}+\sin \left(90^{\circ}-62^{\circ}\right) =\mathrm{k}^{3} \)
\(\cos 28^{\circ}+\cos 62^{\circ} =\mathrm{k}^{3} \)
\(2 \cos 45^{\circ} \cos 17^{\circ} =\mathrm{k}^{3} \)
\(2 \frac{1}{\sqrt{2}} \cos 17^{\circ} =\mathrm{k}^{3} \)
\(\cos 17^{\circ} =\frac{\mathrm{k}^{3}}{\sqrt{2}} \)