In one second it rotates = 2 radians.
For 2 radians it takes 1 second.
\(\text { For } 2 \pi \text { ( } 1 \text { revolution) it will take } \frac{2 \pi}{2}=\pi \text { second. }\)
\(\therefore \text { For } 10 \text { revolutions it takes } 10 \pi \text { seconds. }\)

\(x^{2}+a x+b=0\)
\(\text { Let } \tan \alpha \text { and } \tan \beta \text { be the roots. }\)
\(\tan \alpha+\tan \beta =-a \)
\(\tan \alpha \tan \beta =b \)
\(\frac{\sin (\alpha+\beta)}{\sin \alpha \sin \beta} =\frac{\sin \alpha \cos \beta+\cos \alpha \sin \beta}{\sin \alpha \sin \beta} \)
\(=\cot \beta+\cot \alpha \)
\(=\frac{1}{\tan \alpha}+\frac{1}{\tan \beta} \)
\(=\frac{\tan \alpha+\tan \beta}{\tan \alpha \tan \beta}=\frac{-a}{b} \)

\(\text { Since }|\cos x|<1\)
\(\text { From option (4), }\)
\(\sec \theta=\frac{1}{4}\)
\(\Rightarrow \cos \theta=4 \text { is not possible. }\)

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

\(\cos 1^{\circ}+\cos 2^{\circ}+\cos 3^{\circ}+\ldots \ldots \ldots+\cos 179^{\circ} \)
\(=\left(\cos 1^{\circ}+\cos 179^{\circ}\right)+\left(\cos 2^{\circ}+\cos 178^{\circ}\right)+ ...\)
\(=2 \cos 90^{\circ} \cos 89^{\circ}+2 \cos 90^{\circ} \cos 80^{\circ}+\ldots \ldots . \)
\(=0+0+0 \ldots \ldots=0 \)

sin(-1110°)= - sin (1110°)
= - sin (360 \(\times\) 3 + 30)
= - sin 30°= \(-\frac{1}{2}\)

sin(480°) = sin (360° + 120°) = sin 120°
sin(90° + 30°) = cos 30° = \(\frac{\sqrt{3}}{2}\)\(\frac{\sqrt 3}{2}\)

0

\({1\over a}+{1\over b}\)

480⁰ = 360⁰ + 120⁰
∴ tan(480) = tan(360⁰ + 120⁰)
= tan 120⁰   [∵ 120⁰ lies in the II quadrant of tan is negative]
= tan(180-60⁰)
= -tan 60⁰  [∵ sine is an odd function]
= -\(\sqrt { 3 } \).

We know that sin²x + cos²x = 1
cos x = ±\(\sqrt { 1-sin^{ 2 }x } \)
Since x lies in the III quadrant, cos x is negative and tan x is positive.
∴ cos x = \(-\sqrt { 1-sin^{ 2 }x } =-\sqrt { 1-\frac { 9 }{ 25 } } \)   [∵ sin x = \(\frac { -3 }{ 5 } \)]
\(=-\sqrt{\frac{16}{25}}=\frac{-4}{5}\) and 
\(\tan x=\frac{\sin x}{\cos x}=\frac{-\frac35}{-\frac45}=\frac34\)

LHS = \(cos\left( \frac { \pi }{ 4 } -A \right) cos\left( \frac { \pi }{ 4 } -B \right) -sin\left( \frac { \pi }{ 4 } -A \right) sin\left( \frac { \pi }{ 4 } -B \right) \)
= \(cos\left\{ \left( \frac { \pi }{ 4 } -A \right) \left( \frac { \pi }{ 4 } -B \right) \right\} \)    [cos A cos B - sin A sin B = cos(A + B)
= \(cos\left( \frac { \pi }{ 2 } -A-B \right) \)    where A = \(\frac { \pi }{ 4 } -A,B=\frac { \pi }{ 4 } -B\)
= \(cos\left( \frac { \pi }{ 2 } -(A+B \right) \)
sin(A + B) = RHS

Let \(x=a\sin { \theta } \Rightarrow \frac { x }{ a } =\sin { \theta } \Rightarrow \sin ^{ -1 }{ \left( \frac { x }{ a } \right) } \)
LHS = \(\tan ^{ -1 }{ \left( \frac { a\sin { \theta } }{ \sqrt { { a }^{ 2 }-{ a }^{ 2 }\sin ^{ 2 }{ \theta } } } \right) } =\tan ^{ -1 }{ \left( \frac { a\sin { \theta } }{ a\sqrt { 1-\sin ^{ 2 }{ \theta } } } \right) } \)
= \(\tan ^{ -1 }{ \left( \frac { \sin { \theta } }{ \sqrt { \cos ^{ 2 }{ \theta } } } \right) } \)
= \(\tan ^{ -1 }{ \left( \frac { \sin { \theta } }{ \cos { \theta } } \right) } \)
= \(\tan ^{ -1 }{ \left( \tan { \theta } \right) } =\theta =\sin ^{ -1 }{ \left( \frac { x }{ a } \right) } \)

Let x = cos θ ⇒ θ = cos^-1x
Now, \(2{ sin }^{ -1 }\sqrt { \frac { 1-x }{ 2 } } =2{ sin }^{ -1 }\sqrt { \frac { 1-cos\theta }{ 2 } } \)
= \(2{ sin }^{ -1 }\sqrt { \frac { { 2sin }^{ 2 }\frac { \theta }{ 2 } }{ 2 } } \)
= \(2{ sin }^{ -1 }\left( sin\frac { \theta }{ 2 } \right) \)
= \(2.\left( \frac { \theta }{ 2 } \right) \)
= θ
= cos^-1x
Also \({ 2cos }^{ -1 }\sqrt { \frac { 1+x }{ 2 } } ={ 2cos }^{ -1 }\sqrt { \frac { 1+cos\theta }{ 2 } } \)
= \({ 2cos }^{ -1 }\sqrt { \frac { 2{ cos }^{ 2 }\frac { \theta }{ 2 } }{ 2 } } ={ 2cos }^{ -1 }\left( cos\frac { \theta }{ 2 } \right) \)
= \(2\left( \frac { \theta }{ 2 } \right) =\theta ={ cos }^{ -1 }\left( x \right) \)
From (1) and (2), \({ cos }^{ -1 }x={ 2sin }^{ -1 }\sqrt { \frac { 1-x }{ 2 } } ={ 2cos }^{ -1 }\sqrt { \frac { 1+x }{ 2 } } \)

Given sin 2x + cos x = 0
cos x = -sin 2x
\(cosx=cos\left( \frac { \pi }{ 2 } +2x \right) \)
\(x=2n\pi \pm \left( \frac { \pi }{ 2 } +2x \right) ,n\in z\)
Taking positive sign we get
\(x=2n\pi +\left( \frac { \pi }{ 2 } +2x \right) \)
\(\Rightarrow -x=2n\pi +\frac { \pi }{ 2 } ,n\in z\)
\(\Rightarrow x=2n\pi -\frac { \pi }{ 2 } \) where m = -n∈z
Taking negative sign we get,
\(x=2n\pi -\left( \frac { \pi }{ 2 } +2x \right) \)
\(\Rightarrow 3x=2n\pi -\frac { \pi }{ 2 } \)
\(x=\frac { 2n\pi }{ 3 } -\frac { \pi }{ 6 } ,n\epsilon z\)
Hence \(x=2m\pi -\frac { \pi }{ 2 } \) or \(\frac { 2n\pi }{ 3 } -\frac { \pi }{ 6 } \) where m, n∈z

Let P and Q be the positions of two ships at the end of 3 hours.

Then OP = 3 \(\times\)24 = 72 km and OQ = 3 \(\times\)32 = 96 km.
Using cosine formula in ΔOPQ, we get
PQ² = OP² + OQ² - 2OP.OQ cos∠POQ
= 72² + 96² -2 \(\times\) 72 \(\times\) 96 \(\times\) cos 60°
= 5184 + 9216 - 6912 = 7488
⇒ PQ =\(\sqrt{7488}\) km = 86.533 km

RHS = \(\frac { c-a\cos { B } }{ b-a\cos { C } } \)
= \(\frac { \left( a\cos { B } +b\cos { A } \right) -a\cos { B } }{ \left( a\cos { C } +b\cos { A } \right) -a\cos { C } } \)
= \(\frac { b\cos { A } }{ c\cos { A } } =\frac { b }{ c } \)
By sine formula, \(\frac { a }{ \sin { A } } =\frac { b }{ \sin { B } } =\frac { c }{ \sin { C } } =2A\)
\(\Rightarrow\) a = 2R sin A, b = 2R sin B and c = 2R sin C
Substituting this is (1) we get
RHS = \(\frac { 2R\sin { B } }{ 2R\sin { C } } =\frac { \sin { B } }{ \sin { C } } \) = LHS.
Hence proved.

LHS = \(cosxcos\left( \frac { \pi }{ 3 } -x \right) cos\left( \frac { \pi }{ 3 } +x \right) \)
\(cosx\left[ { cos }^{ 2 }\left( \frac { \pi }{ 3 } \right) -{ sin }^{ 2 }x \right] \)
where \(A=\frac { \pi }{ 3 } +x\) , B = x
\(cosx\left[ { cos }^{ 2 }\frac { \pi }{ 3 } -{ sin }^{ 2 }x \right] \)
= \(cosx\left[ { \left( cos\frac { \pi }{ 3 } \right) }^{ 2 }-{ sin }^{ 2 }x \right] \)
= \(cosx\left[ { \left( \frac { 1 }{ 2 } \right) }^{ 2 }-{ sin }^{ 2 }x \right] \)
= \(cosx\left( \frac { 1 }{ 4 } -{ sin }^{ 2 }x \right) =cosx\left( \frac { 1 }{ 4 } -\left( 1-{ cos }^{ 2 }x \right) \right) \)
= \(cosx\left( \frac { 1 }{ 4 } -1+{ cos }^{ 2 }x \right) \)
= \(cosx\left( \frac { -3 }{ 4 } +{ cos }^{ 2 }x \right) \)
= \(cosx\left( \frac { -3+4{ cos }^{ 2 }x }{ 4 } \right) \)
= \(\frac { 1 }{ 4 } \left( -3cosx+4{ cos }^{ 3 }x \right) \)
= \(\frac { 1 }{ 4 } \left( cos3x \right) \)
= RHS
Hence proved.

Given 3 tanA tan B = 1
\(\Rightarrow \frac { 3\quad sinA\quad sinB }{ cosA\quad cosB } =1\)
\(\Rightarrow \frac { cosAcosB }{ sinAsinB } =\frac { 3 }{ 1 } \)
\(\Rightarrow \frac { cosAcosB+sinAsinB }{ cosAcosB-sinAsinB } =\frac { 3+1 }{ 3-1 } \)
\(\Rightarrow \frac { cos\left( A-B \right) }{ cos\left( A+B \right) } =\frac { 4 }{ 2 } \)
⇒ 2cos(A + B) = cos(A - B)
Hence proved.