We know the sine formula \( \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2 R \)
Now, \( \frac{a-b}{a+b} \cot \frac{C}{2}=\frac{2 R \sin A-2 R \sin B}{2 R \sin A+2 R \sin B} \cot \frac{C}{2} \)
\( =\frac{\sin A-\sin B}{\sin A+\sin B} \cot \frac{C}{2} \)
\(=\frac{2 \cos \frac{A+B}{2} \sin \frac{A-B}{2}}{2 \sin \frac{A+B}{2} \cos \frac{A-B}{2} \cot \frac{C}{2}} \)
\(=\cot \frac{A+B}{2} \tan \frac{A-B}{2} \cot \frac{C}{2} \)
\(=\cot \left(90^{\circ}-\frac{C}{2}\right) \tan \frac{A-B}{2} \cot \frac{C}{2} \)
\(=\tan \frac{C}{2} \tan \frac{A-B}{2} \cot \frac{C}{2}=\tan \frac{A-B}{2} \) 

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.

Given θ + Φ = \(\alpha\) and tan θ = k tan Φ
an θ = k tan Φ
\(\frac { tan\theta }{ tan\phi } =k\Rightarrow \frac { sin\theta cos\phi }{ cos\theta .sin\phi } =\frac { k }{ 1 } \)
⇒ \(\frac { sin\theta cos\phi }{ cos\theta .sin\phi } =\frac { k }{ 1 } \)
(By componendo and dividends)
⇒ ​​​​​\(\frac { sin\theta cos\phi -cos\theta sin\phi }{ cos\theta .sin\phi +cos\theta sin\phi } =\frac { k-1 }{ k+1 } \)
⇒  \(\frac { sin\left( \theta -\phi \right) }{ sin\left( \theta +\phi \right) } =\frac { k-1 }{ k+1 } \)
⇒  \(\frac { sin\left( \theta -\phi \right) }{ sin\alpha } =\frac { k-1 }{ k+1 } \)
⇒  \(sin\left( \theta -\phi \right) =\frac { k-1 }{ k+1 } sin\alpha \)

Let x cosθ = \(y\quad cos\left( θ+2\frac { \pi }{ 3 } \right) =z\quad cos\left( θ+2\frac { \pi }{ 3 } \right) =\lambda \)
\(\frac { \lambda }{ x } =cos\theta ,\frac { \lambda }{ y } =cos\left( θ+2\frac { \pi }{ 3 } \right) \quad \)and \(\frac { \lambda }{ z } =cos\left( θ+4\frac { \pi }{ 3 } \right) \)
\(\frac { \lambda }{ x } +\frac { \lambda }{ y } +\frac { \lambda }{ z } =cos\theta +cos\left( θ+2\frac { \pi }{ 3 } \right) +cos\left( θ+4\frac { \pi }{ 3 } \right) \)
= cos θ + cos(120 + θ) + cos(240 + θ)
= cos θ + cos120 cos θ - sin 120 sin θ + cos 240 cos θ - sin 240 sin θ
= cos θ - cos 60 cos θ - sin 60 sin θ - sin 30 cos θ + cos 30 sin θ
= cos θ - \(\frac{1}{2}cosθ-\frac{\sqrt{3}}{2}sinθ-\frac{1}{2}cosθ+\frac{\sqrt{3}}{2}sinθ\)
= cos θ - \(\frac{1}{2}cosθ-\frac{1}{2}cosθ\)
= cos θ - cos θ = 0
\(\therefore \ \lambda =\left( \frac { 1 }{ x } +\frac { 1 }{ y } +\frac { 1 }{ z } \right) =0\Rightarrow \lambda \left( \frac { yz+xz+xy }{ xyz } \right) =0\)
⇒ xy + yz + zx = 0

LHS = sin² A + sin² B + sin²C
= \(\frac { 1 }{ 2 } \left[ 2{ sin }^{ 2 }A+2{ sin }^{ 2 }B+2{ sin }^{ 2 }C \right] \)
= \(\frac { 1 }{ 2 } \left[ \left( 1-cos2A \right) +\left( 1-cos2B \right) +\left( 1-cos2C \right) \right] \)
= \(\frac { 1 }{ 2 } \left[ 3-\left( cos2A+cos2B+cos2C \right) \right] \)
= \(\frac { 1 }{ 2 } \left[ 3-\left( 2cos\left( \frac { 2A+2B }{ 2 } \right) .cos\left( \frac { 2A-2B }{ 2 } \right) +cos2c \right) \right] \)
= \(\frac { 3 }{ 2 } -\left[ cos\left( A+B \right) cos\left( A-B \right) \right] -\frac { 1 }{ 2 } cos2C\)
= \(\frac { 3 }{ 2 } -\left[ cos\left( 180-C \right) cos\left( A-B \right) \right] -\frac { 1 }{ 2 } \left( 2{ cos }^{ 2 }C-1 \right) \)
= \(\frac { 3 }{ 2 } +\left[ cosC\quad cos\left( A-B \right) \right] -{ cos }^{ 2 }C+\frac { 1 }{ 2 } \)
= \(\frac { 3 }{ 2 } +\frac { 1 }{ 2 } +cosC\left[ cos\left( A-B \right) -cosC \right] \)
= 2 + cosC[cos(A-B) + cos(A+B)]
= 2 + cosC[cos(A-B) + cos(A+B)]
= 2 + cos C.2cos A cos B
= 2 + 2cos A cos B cos C
= RHS
Hence proved.

\(LHS=cot(A+{ 15 }^{ 0 })-tan(A-{ 15 }^{ 0 })\)
\(=\frac { cos(A+15) }{ sin(A+15) } -\frac { sin(A-15) }{ cos(A-15) } \)
\(=\frac { cos(A+15)cos(A-15)-sin(A-15)sin(A+15) }{ sin(A+15).cos(A-15) } \)
\(=\frac { { cos }^{ 2 }A-{ sin }^{ 2 }15\left[ { sin }^{ 2 }A-{ sin }^{ 2 }{ 15 }^{ 0 } \right] }{ \frac { 1 }{ 2 } \left[ sin(A+15+A-15)+sin(A+15-A+15) \right] } \)
\(\left[ \because cos(A+B)cos(A-B)={ cos }^{ 2 }A-{ sin }^{ 2 }Bsin(A+B)sin(A-B)={ sin }^{ 2 }A-{ sin }^{ 2 }B\quad and sinAcosB=\frac { 1 }{ 2 } (sin(A+B)+sin(A-B)) \right] \)
\(=\frac { 2\left[ { cos }^{ 2 }A-{ sin }^{ 2 }15-{ sin }^{ 2 }A+{ sin }^{ 2 }{ 15 }^{ 0 } \right] }{ sin(2A)+sin({ 30 }^{ 0 }) } \quad \left[ \because cos2A={ cos }^{ 2 }A-{ sin }^{ 2 }B \right] \)
\(=\frac { 2\left( { cos }^{ 2 }A-{ sin }^{ 2 }A \right) }{ sin2A+\frac { 1 }{ 2 } } =\frac { 2,cos2A\times 2 }{ 2sin2A+1 } \)
\(=\frac { 4cos2A }{ 1+2sin2A } =RHS\)

Given sin θ + cos θ = m
LHS = cos⁶ θ + sin⁶θ
= (cos²θ)³+ (sin² θ)³
= (cos²θ + sin²θ)(cos⁴θ - cos²θ sin²θ + sin⁴θ)
= 1(cos⁴θ - cos²θ sin²θ + sin⁴θ)
= (cos²θ)²+ (sin²θ)² - cos²θsin²θ
= (cos²θ)²+ (sin²θ)²- cos²θ sin²θ
= 1 - 3 sin²\(\theta\) cos²\(\theta\) ....(1)
RHS = \(\frac { 4-3{ \left( { m }^{ 2 }-1 \right) }^{ 2 } }{ 4 } \)
= \(\frac { 4-3{ \left[ { \left( sin\theta +cos\theta \right) }^{ 2 }-1 \right] }^{ 2 } }{ 4 } =\frac { 4-3\left[ { sin }^{ 2 }\theta +{ cos }^{ 2 }\theta +2sin\theta cos\theta -1 \right] }{ 4 } \)
= \(\frac { 4-12{ sin }^{ 2 }\theta { cos }^{ 2 }\theta }{ 4 } =\frac { 4 }{ 4 } -\frac { 12 }{ 4 } { sin }^{ 2 }\theta { cos }^{ 2 }\theta \)
= 1 - 3 sin²\(\theta\) cos²\(\theta\) ...(2)
From (1) and (2), LHS = RHS

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

sin⁴x - sin²x = 0
sin²x(sin²x - 1) = 0
sin²x = 0 or sin²x = 1
sin x = 0 or sin x =土1
sin x = 0
x = 0, π
sin x = sin(90°)
x = 90° = \(\frac{\pi}{2}\)
sin x = -1
sin x = -sin(\(\frac{\pi}{2}\))
sin x = sin\((\pi+\frac{\pi}{2})\)
sin x = sin(\(\frac{3\pi}{2}\))
x = \(\frac{3\pi}{2}\)
From (1), (2) and (3), the values of x are 0°(\(\frac{\pi}{2}\)),\(\frac{3\pi}{2}\) ,π

Given \(cos\theta =\frac { 1 }{ 2 } \left( a+\frac { 1 }{ a } \right) \)
LHS = cos 3θ = 4cos³θ-3cosθ
= \(4\left[ \frac { 1 }{ 2 } { \left( a+\frac { 1 }{ a } \right) }^{ 3 } \right] -3.\frac { 1 }{ 2 } \left( a+\frac { 1 }{ a } \right) \)
= \(4\left( \frac { 1 }{ 8 } \left( { a }^{ 3 }+{ 3a }^{ 2 }\left( \frac { 1 }{ a } \right) +3a\left( \frac { 1 }{ { a }^{ 2 } } \right) +\left( \frac { 1 }{ { a }^{ 3 } } \right) \right) \right) -\frac { 3 }{ 2 } \left( a+\frac { 1 }{ a } \right) \)
= \(\frac { 1 }{ 2 } \left( { a }^{ 3 }+3a+\frac { 3 }{ a } +\frac { 1 }{ { a }^{ 3 } } \right) -\frac { 3 }{ 2 } \left( a+\frac { 1 }{ a } \right) =\frac { 1 }{ 2 } \left( { a }^{ 3 }+\frac { 1 }{ a } \right) +\frac { 1 }{ 2 } \left( 3a+\frac { 3 }{ a } \right) -\frac { 3 }{ 2 } \left( a+\frac { 1 }{ a } \right) \)
= \(\frac { 1 }{ 2 } \left( { a }^{ 3 }+\frac { 1 }{ { a }^{ 3 } } \right) +\frac { 3 }{ 2 } \left( a+\frac { 1 }{ a } \right) -\frac { 3 }{ 2 } \left( a+\frac { 1 }{ a } \right) \)
\(\frac { 1 }{ 2 } \left( { a }^{ 3 }+\frac { 1 }{ { a }^{ 3 } } \right) \) = RHS
Hence proved.

Since B\(\left( \frac { 5 }{ 7 } ,\frac { 2\sqrt { 6 } }{ 7 } \right) \) is the point on the terminal side of an angle θ

OB² = OA² + AB²
= \((\frac{5}{7})^2+(\frac{2\sqrt{6}}{7})^2=\frac{25}{49}+\frac{24}{49}=\frac{49}{49}=1\)
ஃ ⇒ OB = 1
sin θ = \(\frac{opp}{hyp}=\frac{AB}{OB}=\frac{2\sqrt{6}}{7}\) , cos θ = \(\frac{adj}{hyp}=\frac{OA}{OB}=\frac{5}{7}\)
tan θ = \(\frac{opp}{adj}=\frac{\frac{2\sqrt{6}}{7}}{\frac{5}{7}}=\frac{2\sqrt{6}}{7}\times\frac{7}{5}=\frac{2\sqrt{6}}{5}\)
tan θ = \(\frac{\frac{2\sqrt{6}}{7}}{\frac{5}{7}}=\frac{2\sqrt{6}}{7}\times\frac{7}{5}=\frac{2\sqrt{6}}{5}\)
cot θ = \(\frac{5}{2\sqrt{6}}\)
cosec θ = \(\frac{1}{sinθ}=\frac{7}{2\sqrt{6}}\) ; sec θ = \(\frac{1}{cosθ}=\frac{7}{5}\)
and cot θ = \(\frac{1}{tanθ}=\frac{5}{2\sqrt{6}}\)

Given diameter of the circle is 40 cm
r = 20cm

Let  AB = 20 cm be a chord of the circle
Since OA = OB = AB = 20 cm, ΔAOB is equilateral
ஃ θ = ㄥAOB = 60°= 60 \(\times\) \(\frac{\pi}{180}=\frac{\pi}{3}\) radians
Let I be the length of the minor arc of the chord AB.
Then θ = \(\frac{l}{r}\) ⇒ l = rθ ⇒ l = 20(\(\frac{\pi}{3}\))
l = \(\frac{20\pi}{3}\) cm = 20 \(\times\) \(\frac{22}{7}\times\frac{1}{3}\) = 20.95 cm (app)

Prove that sin (45^o + \(\theta\)) - sin (45^o - \(\theta\)) = \(\sqrt { 2 } \sin { \theta } \)
LHS = sin (45^o + \(\theta\)) - sin (45^o - \(\theta\))
= sin 45 cos \(\theta\) + cos 45 sin \(\theta\) - [sin 45 cos \(\theta\) - cos 45 sin \(\theta\)]
= \(\frac { 1 }{ \sqrt { 2 } } \cos { \theta } +\frac { 1 }{ \sqrt { 2 } } \sin { \theta } \frac { -1 }{ \sqrt { 2 } } \cos { \theta } +\frac { 1 }{ \sqrt { 2 } } \sin { \theta } \)
=  \(\frac { 2\sin { \theta } }{ \sqrt { 2 } } \frac { \sqrt { 2 } }{ \sqrt { 2 } } =\frac { 2\sqrt { 2 } \sin { \theta } }{ 2 } =\sqrt { 2 } \sin { \theta } \) = 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

Squaring and adding ac cos θ  = bc and ac sin θ = ad,
we get a²c² = b²c² + a²d²

sin 4x cos 2x = \(\frac{1}{2}\) [sin (4x + 2x) + sin (4x - 2x)]
= \(\frac{1}{2}\) [sin 6x + sin 2x]

Given sin A = \(\frac{4}{5}\)
cos 2A = 1 - 2 sin² A = 1 - 2 \({ \left( \frac { 4 }{ 5 } \right) }^{ 2 }\)
= 1 - 2 \(( \frac { 16 }{ 25 } )=1-\frac{32}{25}=\frac{25-32}{25}=\frac{-7}{25} \)

tan (45^o + A) = \(\frac { 1+\tan { A } }{ 1-\tan { A } } \)
LHS = tan (45^o + A) = \(\frac { \tan { { 45 }^{ o } } +\tan { A } }{ 1+\tan { { 45 }^{ o } } .A\tan { } } =\frac { 1+\tan { A } }{ 1-\tan { A } } \) = RHS
Hence proved.

150⁰
150⁰ = 150 \(\times\) \(\frac { \pi }{ 180 } =\frac { 5\pi }{ 6 } \)

cos 105^o = cos (60 + 45)
= cos 60 cos 45 - sin 60 sin 45
= \(\frac { 1 }{ 2 } .\frac { 1 }{ \sqrt { 2 } } -\frac { \sqrt { 3 } }{ 2 } .\frac { 1 }{ \sqrt { 2 } } =\frac { 1-\sqrt { 3 } }{ 2\sqrt { 2 } } \)

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

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

\(\text { It lies in } \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}} \text { and } 0+1\)
\(=\sqrt{2} \text { and } 1=[1, \sqrt{2}]\)

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

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

\(\cos 2 \theta \cos 2 \phi+\sin ^{2}(\theta-\phi)-\sin ^{2}(\theta+\phi) \)
\(=\cos 2 \theta \cos 2 \phi+\frac{1}{2}-\frac{1}{2} \cos (2 \theta-2 \phi)-\frac{1}{2}+\frac{1}{2} \cos (2 \theta+2 \phi) \)
\(=\cos 2 \theta \cos 2 \phi+\frac{1}{2}[\cos (2 \theta+2 \phi)-\cos (2 \theta-2 \phi)] \)
\(=\cos 2 \theta \cos 2 \phi-\sin 2 \theta \cos 2 \phi \)
\(=\cos (2 \theta+2 \phi) \)
\(=\cos 2(\theta+\phi) \)

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

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

\(4 \sin ^{2} x+3 \cos ^{2} x+\sin \frac{x}{2}+\cos \frac{x}{2} \)
\(=3\left(\sin ^{2} x+\cos ^{2} x\right)+\sin ^{2} x+\sin \frac{x}{2}+\cos \frac{x}{2} \)
\(=3+\sin ^{2} x+\sin \frac{x}{2}+\cos \frac{x}{2} \)
\(\text { Maximum of } \sin x=1\)
\(\text { Maximum value of } \sin ^{2} x=1\)
\(\text { Maximum value of } \sin \frac{x}{2} \text { occurs where } x=45^{\circ}\)
\(\text { Maximum value is } 3+1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=4+\frac{2}{\sqrt{2}}=4+\sqrt{2}\)