Advertisements
Advertisements
प्रश्न
Prove the following:
`(sin x - sin 3x)/(sin^2 x - cos^2 x) = 2sin x`
Advertisements
उत्तर
We have, L.H.S. = `(sin x - sin 3x)/(sin^2 x - cos^2 x)`
= `(2sin ((x - 3x)/2) cos ((x + 3x)/2))/(-cos^2x - sin^2x)`
= `(2sin (-x) cos (2x))/(-cos2x)`
= `(- 2sin x cos(2x))/(-cos2x)` [∵ cos2x - sin2x = cos 2x]
= `(-2 sinx)/-1` [∵ cos2x - sin2x = cos 2x]
= 2 sin x = R.H.S.
APPEARS IN
संबंधित प्रश्न
Prove that: `sin^2 pi/6 + cos^2 pi/3 - tan^2 pi/4 = -1/2`
Prove the following: `(tan(pi/4 + x))/(tan(pi/4 - x)) = ((1+ tan x)/(1- tan x))^2`
Prove the following:
cos2 2x – cos2 6x = sin 4x sin 8x
Prove the following:
cos 6x = 32 cos6 x – 48 cos4 x + 18 cos2 x – 1
If \[\sin A = \frac{4}{5}\] and \[\cos B = \frac{5}{13}\], where 0 < A, \[B < \frac{\pi}{2}\], find the value of the following:
sin (A − B)
If \[\sin A = \frac{4}{5}\] and \[\cos B = \frac{5}{13}\], where 0 < A, \[B < \frac{\pi}{2}\], find the value of the following:
cos (A − B)
If \[\sin A = \frac{12}{13}\text{ and } \sin B = \frac{4}{5}\], where \[\frac{\pi}{2}\] < A < π and 0 < B < \[\frac{\pi}{2}\], find the following:
sin (A + B)
If \[\sin A = \frac{1}{2}, \cos B = \frac{12}{13}\], where \[\frac{\pi}{2}\]< A < π and \[\frac{3\pi}{2}\] < B < 2π, find tan (A − B).
If \[\sin A = \frac{1}{2}, \cos B = \frac{\sqrt{3}}{2}\], where \[\frac{\pi}{2}\] < A < π and 0 < B < \[\frac{\pi}{2}\], find the following:
tan (A + B)
Evaluate the following:
sin 78° cos 18° − cos 78° sin 18°
Evaluate the following:
cos 80° cos 20° + sin 80° sin 20°
If \[\cos A = - \frac{12}{13}\text{ and }\cot B = \frac{24}{7}\], where A lies in the second quadrant and B in the third quadrant, find the values of the following:
sin (A + B)
Prove that
Prove that
Prove that:
If \[\tan A = \frac{5}{6}\text{ and }\tan B = \frac{1}{11}\], prove that \[A + B = \frac{\pi}{4}\].
Prove that:
\[\cos^2 45^\circ - \sin^2 15^\circ = \frac{\sqrt{3}}{4}\]
Prove that:
\[\frac{\sin \left( A - B \right)}{\cos A \cos B} + \frac{\sin \left( B - C \right)}{\cos B \cos C} + \frac{\sin \left( C - A \right)}{\cos C \cos A} = 0\]
If tan A = x tan B, prove that
\[\frac{\sin \left( A - B \right)}{\sin \left( A + B \right)} = \frac{x - 1}{x + 1}\]
Prove that:
Prove that:
If sin α sin β − cos α cos β + 1 = 0, prove that 1 + cot α tan β = 0.
If \[\tan\theta = \frac{\sin\alpha - \cos\alpha}{\sin\alpha + \cos\alpha}\] , then show that \[\sin\alpha + \cos\alpha = \sqrt{2}\cos\theta\].
Find the maximum and minimum values of each of the following trigonometrical expression:
sin x − cos x + 1
Reduce each of the following expressions to the sine and cosine of a single expression:
\[\sqrt{3} \sin x - \cos x\]
Prove that \[\left( 2\sqrt{3} + 3 \right) \sin x + 2\sqrt{3} \cos x\] lies between \[- \left( 2\sqrt{3} + \sqrt{15} \right) \text{ and } \left( 2\sqrt{3} + \sqrt{15} \right)\]
If x cos θ = y cos \[\left( \theta + \frac{2\pi}{3} \right) = z \cos \left( \theta + \frac{4\pi}{3} \right)\]then write the value of \[\frac{1}{x} + \frac{1}{y} + \frac{1}{z}\]
If A + B + C = π, then \[\frac{\tan A + \tan B + \tan C}{\tan A \tan B \tan C}\] is equal to
If α and β are the solutions of the equation a tan θ + b sec θ = c, then show that tan (α + β) = `(2ac)/(a^2 - c^2)`.
If 3 tan (θ – 15°) = tan (θ + 15°), 0° < θ < 90°, then θ = ______.
If f(x) = cos2x + sec2x, then ______.
[Hint: A.M ≥ G.M.]
If α + β = `pi/4`, then the value of (1 + tan α)(1 + tan β) is ______.
If sinx + cosx = a, then sin6x + cos6x = ______.
If sinx + cosx = a, then |sinx – cosx| = ______.
State whether the statement is True or False? Also give justification.
If cosecx = 1 + cotx then x = 2nπ, 2nπ + `pi/2`
In the following match each item given under the column C1 to its correct answer given under the column C2:
| Column A | Column B |
| (a) sin(x + y) sin(x – y) | (i) cos2x – sin2y |
| (b) cos (x + y) cos (x – y) | (ii) `(1 - tan theta)/(1 + tan theta)` |
| (c) `cot(pi/4 + theta)` | (iii) `(1 + tan theta)/(1 - tan theta)` |
| (d) `tan(pi/4 + theta)` | (iv) sin2x – sin2y |
