Let R₁ and R₂ be two radar stations and A be the position of fighter aircraft at the time of detection.
Let x be the required altitude of the aircraft.

Draw \(\bot AN\) from A to R₁R₂ meeting at N.
\(\angle A=180^0-(30^0+45^0)=105^0\)
Thus, \(\frac{a}{sin\ 45^0}=\frac{100}{sin\ 105^0}\) \(\Rightarrow a=\frac{100}{\frac{\sqrt 3+1}{2\sqrt 2}}\times\frac{1}{\sqrt 2}=\frac{200(\sqrt 3-1)}{2}\)\(=100\times(\sqrt 3-1)km\)
Now, \(sin\ 30^0=\frac{x}{a}\Rightarrow x=50\times(\sqrt 3-1)km\)

​​​​​​The law of sine: \(({a\over sin A})=({b\over sin B})=({c\over sin C})=2R\)
LHS \(={{a^2+b^2}\over{a^2+c^2}}={{(2R \sin A)}^{2}+{(2R \sin B)}^{2}\over{{(2 R \sin A)}^{2}+{(2R \sin C)}^{2}}}\)
\(={{\sin^2 A+\sin^2 B}\over{\sin^2 A+\sin^2 C}}={{1-\cos^2 A+\sin^2 B}\over{1-\cos^2 A+\sin^2 C}}\)
\(={{1-({\cos}^{2}A-{\sin}^{2}B)}\over{1-({\cos}^{2}A-{\sin}^{2}C)}}={1-\cos(A+B)\cos(A-B)\over1-\cos(A+C)\cos(A-C)}\)
\(={1+\cos(A-B)\cos C\over 1+\cos (A-C)\cos B}\)

\(\sqrt{3}\)  sin \(\theta\) - cos \(\theta\) =\(\sqrt{2}\)
Here a = -1; b =\(\sqrt{3};c=\sqrt{2};r=\sqrt{a^2+b^2}=2\)
Thus, the given equation can be written as
\(\sqrt{3}\over2\) sin \(\theta -{1\over 2}cos \theta={1\over\sqrt{2}}\)
\(sin \theta cos {\pi\over6}-cos \theta sin {\pi\over 6}=sin {\pi\over4}\)
\(sin (\theta={\pi\over6})=sin {\pi\over4}\)
\(\theta={\pi\over6}=n\pi\pm(-1)^n {\pi\over4},n \in Z\)
Thus, \(\theta=n\pi\pm{\pi\over6}\pm(-1)^n {\pi\over4},n \in Z\)

i. cos A + cos B + cos C = 2 cos \(({A+B\over 2})\) cos \(({A-B\over 2})\) + cos C
= 2cos \(({\pi \over2}-{c\over2})cos ({A\over2}-{B\over2})+cos C \ (\frac{A+B+C}{2}=\frac{\pi}{2})\)
= 2sin \(({c\over2})cos ({A\over2}-{B\over2})+1-2 sin^2( {C\over 2})\)
= 1 + 2sin \(({c\over2})[cos ({A\over2}-{B\over2})- sin( {C\over 2})]\)
= 1 + 2sin \(({c\over2})[cos ({A\over2}-{B\over2})- cos({\pi\over2}- {C\over 2})]\)
= 1 + 2sin \(({c\over2})[cos ({A\over2}-{B\over2})- cos({A\over2}+ {B\over 2})]\)
= 1 + 4sin \(({A\over 2})sin({B\over2})sin({C\over 2})\)
ii. Let u = sin \(({A\over 2})sin({B\over2})sin({C\over 2})\)
\(=-{1\over2}[cos({A+B\over2})-cos({A-B\over2})]sin{c\over2}\)
\(=-{1\over2}[cos({A+B\over2})-cos({A-B\over2})]cos{A+B\over2}\)
\(=cos^2{A+B\over2}-cos{A-B\over2}cos{A+B\over2}+2u=0,\) which is quadratic in cos \({A+B\over2}\)
Since cos \({A+B\over2}\)  is real number, the above equation has a solution.
Thus, the discriminant b² - 4ac \(\ge\) 0, which gives
\(=cos^2{A-B\over2}-8u \ge 0 \Rightarrow u \le {1\over8}cos^2{A-B\over2}\le{1\over 8}\)
Hence, sin \(({A\over 2})sin({B\over2})sin({C\over 2})\le{1\over 8}\)
iii. From (i) and (ii), we have cos A + cos B + cos C > 1 and cos A + cos B + cos C ≤ 1 + 4 × \({1\over 8}\)
Thus, we get 1 < cos A + cos B + cos C \(\le\frac{3}{2}\)

Using sine formula,
Let \(\frac { sinA }{ a } =\frac { sinB }{ b } =\frac { sinC }{ c } =k\)
sin A = ak, sin B = bk, sin C = ck
Given \(\frac { sin\quad A }{ sin\quad C } =\frac { sin\left( A-B \right) }{ sin\left( B-C \right) } \)
\(\Rightarrow \frac { sin\left( B+C \right) }{ sin\left( A+B \right) } =\frac { sin\left( A-B \right) }{ sin\left( B-C \right) } \)
⇒ Sin(B + C) sin(B - C) = sin(A + B) sin(A - B)
⇒ sin²B - sin²C = sin²A - sin²B
⇒ k²b²- k²c² = k²a² - k²b²
⇒ b²- c² = a²- b²
⇒ 2b² = a² + c²
⇒ a², b², c² are in A.P.
Hence proved

Let \(\frac { a }{ sinA } =\frac { b }{ sinB } =\frac { c }{ sinC } =k\)
a = k sin A, b = k sinB, c = k sin C...(1)
Consider \(\frac { asin\left( B-C \right) }{ { b }^{ 2 }-{ c }^{ 2 } } =\frac { ksinAsin\left( B-C \right) }{ { k }^{ 2 }{ sin }^{ 2 }B-{ k }^{ 2 }{ sin }^{ 2 }C } \)
\(\frac { ksin\left( B+C \right) sin\left( B-C \right) }{ { k }^{ 2 }\left( { sin }^{ 2 }B-{ sin }^{ 2 }C \right) } =\frac { k\left( { sin }^{ 2 }B-{ sin }^{ 2 }C \right) }{ { k }^{ 2 }\left( { sin }^{ 2 }B-{ sin }^{ 2 }C \right) } =\frac { 1 }{ k } \)...(2)
\(\frac { bsin\left( C-A \right) }{ { c }^{ 2 }-{ a }^{ 2 } } =\frac { ksinB\quad sin\left( C-A \right) }{ { k }^{ 2 }{ sin }^{ 2 }C-{ k }^{ 2 }{ sin }^{ 2 }A } =\frac { sin\left( C+A \right) \quad sin\left( C-A \right) }{ ksin\left( C+A \right) \quad sin\left( C-A \right) } =\frac { 1 }{ k } \)...(3)
Similarly, \(\frac { Csin\left( A-B \right) }{ { a }^{ 2 }-{ b }^{ 2 } } =\frac { 1 }{ k } \)...(4)
From (2), (3) and (4)
\(\frac { asin\left( B-C \right) }{ { b }^{ 2 }-{ c }^{ 2 } } =\frac { bsin\left( C-A \right) }{ { c }^{ 2 }-{ a }^{ 2 } } =\frac { csin\left( A-B \right) }{ { a }^{ 2 }-{ b }^{ 2 } } \)

Consider \(\frac { b+c }{ a } =\frac { ksinB+ksinC }{ ksinA } \)
\(\Rightarrow \frac { sinB+sinC }{ sinA } =\frac { 2sin\left( \frac { B+C }{ 2 } \right) cos\left( \frac { B-C }{ 2 } \right) }{ 2sin\frac { A }{ 2 } cos\frac { A }{ 2 } } \)
\(\Rightarrow \frac { sin\left( 90-\frac { A }{ 2 } \right) cos\left( \frac { B-C }{ 2 } \right) }{ sin\frac { A }{ 2 } cos\frac { A }{ 2 } } \)
\(\Rightarrow \frac { cos\frac { A }{ 2 } cos\left( \frac { B-C }{ 2 } \right) }{ sin\frac { A }{ 2 } cos\frac { A }{ 2 } } =\frac { cos\left( \frac { B-C }{ 2 } \right) }{ sin\frac { A }{ 2 } } \)
\(\Rightarrow \frac { cos\left( \frac { B-\left( 180-A-B \right) }{ 2 } \right) }{ sin\frac { A }{ 2 } } \)
\(\Rightarrow \frac { cos\left( -90+\frac { A }{ 2 } +B \right) }{ sin\frac { A }{ 2 } } =\frac { cos\left[ -\left( 90-\left( \frac { A }{ 2 } +B \right) \right) \right] }{ sin\frac { A }{ 2 } } \)
\(\Rightarrow \frac { cos\left( 90-\left( \frac { A }{ 2 } +B \right) \right) }{ sin\frac { A }{ 2 } } \)
\(\frac { b+c }{ a } \Rightarrow \frac { sin\left( \frac { A }{ 2 } +B \right) }{ sin\frac { A }{ 2 } } \)
\(\Rightarrow \left( b+c \right) sin\frac { A }{ 2 } =asin\left( \frac { A }{ 2 } +B \right) \)
Hence proved

We have A + B + C = 180°
B + C = 180° - A = 180° - 60° = 120°
\(\Rightarrow \frac { B+C }{ 2 } =60°\)...(1)
Using sine formula, \(\frac { a }{ sinA } =\frac { b }{ sinB } =\frac { c }{ sinC } =k\)
a = k sin A, b = k sin B, c = k sin C
RHS = 2a cos\(\left( \frac { B-C }{ 2 } \right) \)...(2)
= 2.K sin A cos\(\left( \frac { B-C }{ 2 } \right) \)
=2.K sin 60° cos \(\left( \frac { B-C }{ 2 } \right) \)
= 2K.sin\(\left( \frac { B+C }{ 2 } \right) \)cos\(\left( \frac { B-C }{ 2 } \right) \)
= K\(\left[ sin\left( \frac { B+C+B-C }{ 2 } \right) +sin\left( \frac { B+C-B-C }{ 2 } \right) \right] \)
= K[sin B + sin C] 
= K sin B + K sin C
= b + c [From(2)]
LHS Hence proved.

LHS = \(\frac { cot(180+\theta )sin(90-\theta )cos(-\theta ) }{ sin(270+\theta )tan(-\theta )cosec(360+\theta ) } \)
cot (180 + \(\theta \)) = cot \(\theta \)  sin (270+\(\theta \)) = -cos \(\theta \)
sin (90 - \(\theta \) ) = cos \(\theta \)  tan(-\(\theta \)) = -tan \(\theta \)
cos (-\(\theta \)) = cos \(\theta \)  cosec (360 +\(\theta \)) = cosec \(\theta \)
\(\frac { cot\theta .cos\theta .cos\theta }{ (-cos\theta )(-tan\theta )(cosec\theta ) } =\frac { \frac { cos\theta }{ sin\theta } .cos\theta .cos\theta }{ cos\theta .\frac { sin\theta }{ cos\theta } .\frac { 1 }{ sin\theta } } \)
\(=\frac { { cos }^{ 3 }\theta }{ sin\theta } ={ cos }^{ 2 }\theta .\frac { cos\theta }{ sin\theta } \)
\(={ cos }^{ 2 }\theta .cot\theta \)

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.

Given \(\sin ^{ 2 }{ \theta } =\frac { 3 }{ 5 } \)
\(\Rightarrow \sin { \theta } =\pm \frac { \sqrt { 3 } }{ 2 } \)
\(\Rightarrow \sin { \theta } =\frac { \sqrt { 3 } }{ 2 } or\quad \sin { \theta } =\frac { -\sqrt { 3 } }{ 2 } \)
\(\Rightarrow \sin { \theta } =\sin { { 60 }^{ o } } \quad or\quad \sin { \theta } =-\sin { { 60 }^{ o } } \)
\(\Rightarrow \sin { \theta } =\sin { { 60 }^{ o } } \quad or\quad \theta =180-{ 60 }^{ o }\)
\(\\ \Rightarrow \theta ={ 60 }^{ o }or\quad \theta ={ 120 }^{ o }\)

Given cos θ = -\(\frac{1}{2}\) , θ lies in the III quadrant
    
AB² = AC² - BC²
= 4 - 1 = 3
AB = √3
Since θ lies in the III quadrant, only tan θ and cot θ are positive.
Sin θ = \(\frac{opp}{hyp}\) = -\(\frac{\sqrt{3}}{2}\) ; and tan θ = √3
Cosec θ = \(\frac{1}{sin θ}=\frac{-2}{\sqrt{3}}\) ; sec θ = \(\frac{1}{cos θ}\) = -2; cot θ = \(\frac{1}{tan θ}=\frac{1}{\sqrt{3}}\)

\(LHS=\frac { sin4x+sin2x }{ cos4x+cos2x } \)
\(=\frac { 2sin\left( \frac { 4x+2x }{ 2 } \right) .cos\left( \frac { 4x-2x }{ 2 } \right) }{ 2cos\left( \frac { 4x+2x }{ 2 } \right) .cos\left( \frac { 4x-2x }{ 2 } \right) } =\frac { sin3x.cosx }{ cos3x.cosx } =tan3x\) = RHS

\(LHS=\frac { (cos\theta -cos3\theta )(sin8\theta +sin2\theta ) }{ (sin5\theta -sin\theta )(cos4\theta -cos6\theta ) } \)
\(=\frac { 2sin\left( \frac { \theta +3\theta }{ 2 } \right) sin\left( \frac { 3\theta -\theta }{ 2 } \right) .2sin\left( \frac { 8\theta +2\theta }{ 2 } \right) cos\left( \frac { 8\theta -2\theta }{ 2 } \right) }{ 2cos\left( \frac { 5\theta +\theta }{ 2 } \right) sin\left( \frac { 5\theta -\theta }{ 2 } \right) .2sin\left( \frac { 4\theta +6\theta }{ 2 } \right) sin\left( \frac { 6\theta -4\theta }{ 2 } \right) } \)
\(=\frac { sin2\theta .sin\theta .sin5\theta .cos3\theta }{ cos3\theta .sin2\theta .sin5\theta .sin\theta } =1=RHS\)

\(LHS=\frac { sin8x\quad cosx-sin6x\quad cos3x }{ cos2x\quad cosx-sin3x\quad sin4x } \)
\(=\frac { \frac { 1 }{ 2 } \left[ sin9x+sin(7x) \right] -\frac { 1 }{ 2 } \left[ sin9x+sin3x \right] }{ \frac { 1 }{ 2 } \left[ cos\quad 3x+cos\quad x) \right] -\frac { 1 }{ 2 } \left[ cos\quad (x)-cos7x) \right] } \quad \left[ \therefore sinAcosB=\frac { 1 }{ 2 } (sin(A+B)+sin(A-B)) \right] \)
\(=\frac { \frac { 1 }{ 2 } \left[ sin9x+sin7x-sin9x-sin3x \right] }{ \frac { 1 }{ 2 } \left[ cos3x+cosx-cosx+cos7x \right] } =\frac { sin7x-sin3x }{ cos3x+cos7x } \)
\(=\frac { 2cos\left( \frac { 7x+3x }{ 2 } \right) sin\left( \frac { 7x-3x }{ 2 } \right) }{ 2cos\left( \frac { 7x+3x }{ 2 } \right) .cos\left( \frac { 7x-3x }{ 2 } \right) } =\frac { 2cos5x.sin2x }{ 2cos5x.cos2x } \)
\(\left[ \because sinC-sinD=2cos\left( \frac { C+D }{ 2 } \right) sins\left( \frac { C-D }{ 2 } \right) andcosC+cosD=2cos\left( \frac { C+D }{ 2 } \right) coss\left( \frac { C-D }{ 2 } \right) \right] \)
= tan 2x = RHS