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PUC Science कक्षा ११ - Karnataka Board PUC Question Bank Solutions for Mathematics

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Mathematics
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If \[a \cos2x + b \sin2x = c\]  has α and β as its roots, then prove that

(ii)  \[\tan\alpha \tan\beta = \frac{c - a}{c + a}\]

 

[3] Trigonometric Functions
Chapter: [3] Trigonometric Functions
Concept: undefined >> undefined

If \[a \cos2x + b \sin2x = c\]  has α and β as its roots, then prove that

(iii)\[\tan\left( \alpha + \beta \right) = \frac{b}{a}\] 

 

[3] Trigonometric Functions
Chapter: [3] Trigonometric Functions
Concept: undefined >> undefined

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If  \[\cos\alpha + \cos\beta = 0 = \sin\alpha + \sin\beta\] , then prove that \[\cos2\alpha + \cos2\beta = - 2\cos\left( \alpha + \beta \right)\] .

 
[3] Trigonometric Functions
Chapter: [3] Trigonometric Functions
Concept: undefined >> undefined

Prove that:  \[\sin 5x = 5 \sin x - 20 \sin^3 x + 16 \sin^5 x\]

 
[3] Trigonometric Functions
Chapter: [3] Trigonometric Functions
Concept: undefined >> undefined

Prove that: \[4 \left( \cos^3 10 °+ \sin^3 20° \right) = 3 \left( \cos 10°+ \sin 2° \right)\]

 
[3] Trigonometric Functions
Chapter: [3] Trigonometric Functions
Concept: undefined >> undefined

Prove that:  \[\cos^3 x \sin 3x + \sin^3 x \cos 3x = \frac{3}{4} \sin 4x\]

 
[3] Trigonometric Functions
Chapter: [3] Trigonometric Functions
Concept: undefined >> undefined

Prove that `tan x + tan (π/3 + x) - tan(π/3 - x) = 3tan 3x`

[3] Trigonometric Functions
Chapter: [3] Trigonometric Functions
Concept: undefined >> undefined

\[\tan x + \tan\left( \frac{\pi}{3} + x \right) - \tan\left( \frac{\pi}{3} - x \right) = 3 \tan 3x\] 

[3] Trigonometric Functions
Chapter: [3] Trigonometric Functions
Concept: undefined >> undefined
\[\cot x + \cot\left( \frac{\pi}{3} + x \right) + \cot\left( \frac{\pi}{3} - x \right) = 3 \cot 3x\]

 

[3] Trigonometric Functions
Chapter: [3] Trigonometric Functions
Concept: undefined >> undefined

\[\cot x + \cot\left( \frac{\pi}{3} + x \right) + \cot\left( \frac{2\pi}{3} + x \right) = 3 \cot 3x\] 

[3] Trigonometric Functions
Chapter: [3] Trigonometric Functions
Concept: undefined >> undefined
\[\sin 5x = 5 \cos^4 x \sin x - 10 \cos^2 x \sin^3 x + \sin^5 x\]

 

[3] Trigonometric Functions
Chapter: [3] Trigonometric Functions
Concept: undefined >> undefined
\[\sin^3 x + \sin^3 \left( \frac{2\pi}{3} + x \right) + \sin^3 \left( \frac{4\pi}{3} + x \right) = - \frac{3}{4} \sin 3x\]

 

[3] Trigonometric Functions
Chapter: [3] Trigonometric Functions
Concept: undefined >> undefined

Prove that \[\left| \sin x \sin \left( \frac{\pi}{3} - x \right) \sin \left( \frac{\pi}{3} + x \right) \right| \leq \frac{1}{4}\]  for all values of x

 
 
[3] Trigonometric Functions
Chapter: [3] Trigonometric Functions
Concept: undefined >> undefined

Prove that \[\left| \cos x \cos \left( \frac{\pi}{3} - x \right) \cos \left( \frac{\pi}{3} + x \right) \right| \leq \frac{1}{4}\]  for all values of x

 
[3] Trigonometric Functions
Chapter: [3] Trigonometric Functions
Concept: undefined >> undefined

Prove that: \[\sin^2 \frac{2\pi}{5} - \sin^{2 -} \frac{\pi}{3} = \frac{\sqrt{5} - 1}{8}\]

  
[3] Trigonometric Functions
Chapter: [3] Trigonometric Functions
Concept: undefined >> undefined

Prove that: \[\sin^2 24°- \sin^2 6° = \frac{\sqrt{5} - 1}{8}\]

  
[3] Trigonometric Functions
Chapter: [3] Trigonometric Functions
Concept: undefined >> undefined

Prove that:  \[\sin^2 42° - \cos^2 78 = \frac{\sqrt{5} + 1}{8}\] 

 
[3] Trigonometric Functions
Chapter: [3] Trigonometric Functions
Concept: undefined >> undefined

Prove that:  \[\cos 78°  \cos 42°  \cos 36° = \frac{1}{8}\]

[3] Trigonometric Functions
Chapter: [3] Trigonometric Functions
Concept: undefined >> undefined

Prove that: \[\cos\frac{\pi}{15}\cos\frac{2\pi}{15}\cos\frac{4\pi}{15}\cos\frac{7\pi}{15} = \frac{1}{16}\]

 
[3] Trigonometric Functions
Chapter: [3] Trigonometric Functions
Concept: undefined >> undefined

Prove that: \[\cos 6° \cos 42°   \cos 66°    \cos 78° = \frac{1}{16}\]

 
[3] Trigonometric Functions
Chapter: [3] Trigonometric Functions
Concept: undefined >> undefined
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