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

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Mathematics
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In a ∆ABC, prove that:

\[\cos\left( \frac{A + B}{2} \right) = \sin\frac{C}{2}\]

 

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

In a ∆ABC, prove that:

\[\tan\frac{A + B}{2} = \cot\frac{C}{2}\]
[3] Trigonometric Functions
Chapter: [3] Trigonometric Functions
Concept: undefined >> undefined

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In a ∆A, B, C, D be the angles of a cyclic quadrilateral, taken in order, prove that cos(180° − A) + cos (180° + B) + cos (180° + C) − sin (90° + D) = 0

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

Find x from the following equations:
\[cosec\left( \frac{\pi}{2} + \theta \right) + x \cos \theta \cot\left( \frac{\pi}{2} + \theta \right) = \sin\left( \frac{\pi}{2} + \theta \right)\]

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

Find x from the following equations:
\[x \cot\left( \frac{\pi}{2} + \theta \right) + \tan\left( \frac{\pi}{2} + \theta \right)\sin \theta + cosec\left( \frac{\pi}{2} + \theta \right) = 0\]

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

Prove that:
\[\tan 4\pi - \cos\frac{3\pi}{2} - \sin\frac{5\pi}{6}\cos\frac{2\pi}{3} = \frac{1}{4}\]

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

Prove that:
\[\sin\frac{13\pi}{3}\sin\frac{8\pi}{3} + \cos\frac{2\pi}{3}\sin\frac{5\pi}{6} = \frac{1}{2}\]

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

Prove that:
\[\sin \frac{13\pi}{3}\sin\frac{2\pi}{3} + \cos\frac{4\pi}{3}\sin\frac{13\pi}{6} = \frac{1}{2}\]

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

Prove that:

\[\sin\frac{10\pi}{3}\cos\frac{13\pi}{6} + \cos\frac{8\pi}{3}\sin\frac{5\pi}{6} = - 1\]
[3] Trigonometric Functions
Chapter: [3] Trigonometric Functions
Concept: undefined >> undefined

Prove that:

\[\tan\frac{5\pi}{4}\cot\frac{9\pi}{4} + \tan\frac{17\pi}{4}\cot\frac{15\pi}{4} = 0\]

 

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

If tan x = \[x - \frac{1}{4x}\], then sec x − tan x is equal to

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

If sec \[x = x + \frac{1}{4x}\], then sec x + tan x = 

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

If \[\frac{\pi}{2} < x < \frac{3\pi}{2},\text{ then }\sqrt{\frac{1 - \sin x}{1 + \sin x}}\] is equal to

 

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

Using binomial theorem, indicate which is larger (1.1)10000 or 1000.

 
[7] Binomial Theorem
Chapter: [7] Binomial Theorem
Concept: undefined >> undefined
\[\sqrt{\frac{1 + \cos x}{1 - \cos x}}\] is equal to

 

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

If \[0 < x < \frac{\pi}{2}\], and if \[\frac{y + 1}{1 - y} = \sqrt{\frac{1 + \sin x}{1 - \sin x}}\], then y is equal to

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

If \[\frac{\pi}{2} < x < \pi, \text{ then }\sqrt{\frac{1 - \sin x}{1 + \sin x}} + \sqrt{\frac{1 + \sin x}{1 - \sin x}}\] is equal to

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

If x = r sin θ cos ϕ, y = r sin θ sin ϕ and r cos θ, then x2 + y2 + z2 is independent of

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

If tan x + sec x = \[\sqrt{3}\], 0 < x < π, then x is equal to

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

If tan \[x = - \frac{1}{\sqrt{5}}\] and θ lies in the IV quadrant, then the value of cos x is

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