Given sin 2x + sin 4x + sin 6x = 0
⇒ sin 4x + (sin 2x + sin 6x) = 0
\(\Rightarrow sin4x+2sin\left( \frac { 6x+2x }{ 2 } \right) cos\left( \frac { 6x-2x }{ 2 } \right) =0\)
⇒ sin 4x + 2 sin 4x . cos 2x = 0
⇒ sin 4x (1 + 2 cos 2x) = 0
⇒ sin 4x = 0 or 1 + 2 cos 2x = 0
⇒ sin 4x = 0, cos 2x = \(\frac{-1}{2}\)
case (i) : 
When sin 4x = 0
⇒ 4x = nπ, n∈z
⇒ x = \(\frac { n\pi }{ 4 } ,\ n\epsilon z\)
case(ii) :
When cos 2x = \(\frac{-1}{2}\)
\(\Rightarrow cos2x=-cos\left( \frac { \pi }{ 3 } \right) \)
\(cos2x=-cos\left( \pi -\frac { \pi }{ 3 } \right) \)
\(cos2x=-cos\frac { 2\pi }{ 3 } m\epsilon z\)
\(x=m\pi \pm \frac { \pi }{ 3 } ,\quad m\epsilon z\)
Hence \(x=\frac { n\pi }{ 4 } \) or x = \(m\pi \pm \frac { \pi }{ 3 } \), where m,n∈z

We have
AB² = AC² + BC² - 2AC . BC.cos\(\frac{\pi}{4}\)

⇒ AB² = 250² + 3002 - 2 . 250 \(\times\) 300 \(\times\) \(\frac{1}{\sqrt{2}}\)
⇒ AB² = 62500 + 90000 - 75000√2
⇒ AB² = 152500 - 75000 \(\times\) 1.44
⇒ AB² = 44500
⇒ AB = \(\sqrt{44500}\) = 210.95 m

Given a cos (x + y) = b cos (x - y)
\(\Rightarrow\) a [cos x cos y - sin x sin y] = b [cos x cos y + sin x sin y]
\(\Rightarrow\) a cos x cos y - a sin x sin y = b cos x cos y + b sin x sin y
\(\Rightarrow\) (a - b) cos x cos y = (a + b) sin x sin y
\(\Rightarrow\) (a - b) \(​​​​\frac{\cos { y } }{\sin { y } } \) = (a + b) \(​​​​\frac {\sin { x } }{\cos { x } } \)
\(\Rightarrow\) (a - b) cot y = (a + b) tan x = Hence proved.

Given cot θ(1 + sin θ) = 4m and cot θ(1-sin θ) = 4n
cot θ + cot θ sin θ = 4m and cot θ - cos θ sin θ = 4n
⇒ cot θ + \(\frac{cosθ}{sinθ}\)sin θ = 4m and cot θ - \(\frac{cosθ}{sinθ}\)sin θ = 4n
⇒ cot θ + cot θ = 4m...(1)
and cot θ - cos θ = 4m...(2)
Squaring and subtracting (1) and (2) we get,
⇒ (cot θ + cos θ)²- (cot θ - cos θ)² = 16m²-16n²
⇒ cot²θ + cos²θ + 2cot θ cos θ - cot²θ - cos²θ + 2cot θ cos θ = 16(m²-n²)
4cot θ cos θ = 16(m²- n²)
⇒ \(\frac{cosθ}{sinθ}\)cos θ = 4(m²- n²)
⇒ \(\frac{cos^2θ}{sinθ}\) = 4(m²- n²)
Squaring both sides we get,
⇒ \(\frac{cos^4θ}{sin^2θ}\) = 16(m²-n²)...(3)
Multiplying (1) and (2) we get,
(cot θ + cos θ)(cot θ - cos θ) = 16mn
cot²θ - cos²θ = 16mn
⇒ \(\frac{cos^2θ}{sin^2θ}\) - cos² θ = 16mn
⇒ ​​​​​​​\(\frac{cos^2θ-cos^2θsin^2θ}{sin^2θ}=16mn\)
⇒ ​​​​​​​\(\frac{cos^2θ(1-sin^2θ)}{sin^2θ}=16mn\)
⇒ \(\frac{cos^2\theta.cos^2\theta}{sin^2\theta}=16mn\)
⇒ \(\frac{cos^4\theta}{sin^2\theta}=16mn\) ...(4)
From (3) and (4)
⇒ 16(m²- n²)² = 16mn
⇒ (m²- n²) = mn

LHS = \(\sin ^{ 2 }{ \frac { \pi }{ 18 } } +\sin ^{ 2 }{ \frac { \pi }{ 9 } } +\sin ^{ 2 }{ \frac { 7\pi }{ 18 } } +\sin ^{ 2 }{ \frac { 4\pi }{ 9 } } \)
\(=​​​​\sin ^{ 2 }{ \left( \frac { \pi }{ 18 } \times \frac { 180 }{ \pi } \right) } +\sin ^{ 2 }{ \left( \frac { \pi }{ 9 } \times \frac { 180 }{ \pi } \right) } +\sin ^{ 2 }{ \left( \frac { 7\pi }{ 18 } \times \frac { 180 }{ \pi } \right) } +\sin ^{ 2 }{ \left( \frac { 4\pi }{ 9 } \times \frac { 180 }{ \pi } \right) } \)
= sin² 10^o + sin² 20^o + sin² 70^o + sin² 80^o
= [sin (90 - 80^o)]² + [sin (90 - 70)]² + sin² 70^o + sin² 80^o
= cos² 80^o + cos² 70^o + sin² 70^o +sin² 80^o
= (cos² 80^o + sin² 80^o) + (cos² 70 + sin² 70)
= 1 + 1 = 2 = RHS
Hence proved.

LHS = 4 sin A sin (60° + A) .sin (60° - A)
= 4 sinA{sin (60° + A). sin (60° -A)
= 4 sin A {sin² 60° - sin² A}
= \(4sinA\left\{\frac{3}{4}-sin^2A\right\}\)= 3 sin A - 4sin³A = sin³A = RHS

cos (90°+ \(\theta\)) = -sin\(\theta\)
sec(-\(\theta\)) = sec \(\theta\)
tan(180°-\(\theta\)) = -tan\(\theta\)
sec(360°-\(\theta\)) = sec\(\theta\)
sin(180°+\(\theta\)) = -sin\(\theta\)
cot(90°+\(\theta\)) = -tan\(\theta\)
\(\therefore\frac{cos(90°+\theta)sec(-\theta)tan(180°-\theta)}{sec(360°-\theta)sin(180°+\theta)cot(90°+\theta)}\)
\(=\frac{(-sin\theta)(sec\theta)(-tan\theta)}{(sec\theta)(-sin\theta)(-tan\theta)}=1\)

LHS = 1 + cos 2x + cos 4x + cos 6x
= ( cos 0x + cos 2x ) + ( cos 4x + cos 6x )
\(=\left[ 2 cos \left( {{0+2}\over{2}} \right)x.cos \left({{2x-x}\over{2}} \right)\right]+\left[ 2\ cos\left({{4x+6x}\over{2}} \right).cos\left( {{6x-4x}\over{2}} \right) \right]\left[ \because cos\ C+cos\ D=2cos\left( {{C+D}\over{2}} \right) .cos\left( {{C-D}\over{2}} \right)\right]\)
= 2 cos x. cos x + 2 cos 5x. cos x
= 2 cos x ( cos x + cos 5x )
= 2 cos x . \(\left( 2\ cos\left({{5x+x}\over{2}} \right) .cos\left( {{5x-x}\over{2}} \right)\right) \)
= 2 cos x.2.cos 3x. cos 2x
= 4 cos x cos 2x cos 3x = RHS.
Hence proved.

LHS = sin x + sin 2x + sin 3x
= (sin x + sin 3x) + sin 2x
\(=2sin\left( \frac { x+3x }{ 2 } \right) cos\left( \frac { x-3x }{ 2 } \right) +sin2x\)
= 2sin 2x.cos(-x) + sin 2x
= 2sin 2x + cos x + sin 2x
= sin 2x(1 + 2cos x) = RHS

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
\(sinA=\frac { 3 }{ 5 } \quad sinB=\frac { 40 }{ 41 } \)
\(cosA=\frac { 4 }{ 5 } \quad cosB=\frac { 9 }{ 41 } \)
sin (A + B) = sinA cosB + cosA sinB
=\(\frac { 3 }{ 5 } .\frac { 9 }{ 41 } +\frac { 4 }{ 5 } .\frac { 40 }{ 41 } \)
\(=\frac { 27 }{ 205 } +\frac { 160 }{ 205 } =\frac { 187 }{ 205 } \)

\({ tan }^{ -1 }\left( \frac { 1 }{ 7 } \right) +{ tan }^{ -1 }\left( \frac { 1 }{ 13 } \right) ={ tan }^{ -1 }\left( \frac { \frac { 1 }{ 7 } +\frac { 1 }{ 13 } }{ 1-\frac { 1 }{ 7 } \frac { 1 }{ 13 } } \right) ={ tan }^{ -1 }\left( \frac { 20 }{ 90 } \right) ={ tan }^{ -1 }\frac { 2 }{ 9 } \)

Let \(\left( cos^{ -1 }\left( \frac { 3 }{ 5 } \right) \right) =\theta \)
∴ \(\theta \in \left[ 0,\frac { \pi }{ 2 } \right] \quad cos\theta =\frac { 3 }{ 5 } \)
⇒ \(sin\theta =+\sqrt { 1-cos^{ 2 }\theta } =\sqrt { 1-\frac { 9 }{ 25 } } =\sqrt { \frac { 16 }{ 25 } } =\frac { 4 }{ 5 } \)
∴ \(sin\left( cos^{ -1 }\left( \frac { 3 }{ 5 } \right) \right) =\frac { 4 }{ 5 } \)

\(\frac { cos(2\pi +x)cosec(2\pi +x)tan\left( \frac { \pi }{ 2 } +x \right) }{ sec\left( \frac { \pi }{ 2 } +x \right) cos.cot(\pi +x) } \) = 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

sin\(\left( \frac { -11\pi }{ 3 } \right) =-sin\frac { 11\pi }{ 3 } \)
= \(-sin\left( \frac { 11\times 180 }{ 3 } \right) \) = -sin(660⁰)
= -sin(2 \(\times\) 360⁰ - 60⁰)
= -(-sin(60⁰))   [Angle is in the IV quadrant and sine is negative
= sin 60⁰ = \(\frac { \sqrt { 3 } }{ 2 } \).

Let \(sin^{ -1 }\left( \frac { 1 }{ \sqrt { 2 } } \right) \) = y, where \(-\frac { \pi }{ 2 } \le y\le \frac { \pi }{ 2 } \)
⇒ sin y = \(\frac { 1 }{ \sqrt { 2 } } \)
⇒ sin y = sin \(\frac { \pi }{ 4 } \)
⇒ y = \(\frac { \pi }{ 4 } \)
Thus the principal value of \(sin^{ -1 }\left( \frac { 1 }{ \sqrt { 2 } } \right) \)= \(\frac { \pi }{ 4 } \) .

\(\frac{\sin (A-B)}{\cos A \cos B} =\frac{\sin A \cos B-\cos A \sin B}{\cos A \cos B} \)
\(=\tan A-\tan B \)
\(\text { L.H.S } =\tan A-\tan B+\tan B-\tan C+\tan C-\tan A=0\)

\(\frac{\tan 140^{\circ}-\tan 130^{\circ}}{1+\tan 140^{\circ} \tan 130^{\circ}} =\tan \left(140^{\circ}-130^{\circ}\right) \)
\(=\tan 10^{\circ} \ldots \ldots(1) \)
\(\tan 40^{\circ} =\lambda \)
\(\tan 80^{\circ}=\frac{2 \tan 40^{\circ}}{1-\tan ^{2} 40^{\circ}} =\frac{2 \lambda}{1-\lambda^{2}} \)
\(\tan \left(90^{\circ}-10^{\circ}\right) =\frac{2 \lambda}{1-\lambda^{2}} \)
\(\cot 10^{\circ} =\frac{2 \lambda}{1-\lambda^{2}} \)
\(\Rightarrow \tan 10^{\circ} =\frac{1-\lambda^{2}}{2 \lambda} \)
\(\frac{\tan 140^{\circ}-\tan 130^{\circ}}{1+\tan 140^{\circ} \tan 130^{\circ}} \)
\(=\frac{1-\lambda^{2}}{2 \lambda}\)

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