Advertisements
Advertisements
Question
Prove the following:
`(sin x - sin 3x)/(sin^2 x - cos^2 x) = 2sin x`
Advertisements
Solution
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
RELATED QUESTIONS
Prove that `cot^2 pi/6 + cosec (5pi)/6 + 3 tan^2 pi/6 = 6`
Prove the following:
sin2 6x – sin2 4x = sin 2x sin 10x
Prove the following:
`(cos9x - cos5x)/(sin17x - sin 3x) = - (sin2x)/(cos 10x)`
Prove the following:
`(sin x - siny)/(cos x + cos y)= tan (x -y)/2`
Prove the following:
`(sin x + sin 3x)/(cos x + cos 3x) = tan 2x`
Prove that: sin 3x + sin 2x – sin x = 4sin x `cos x/2 cos (3x)/2`
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:
tan (A + B)
Prove that:
Prove that \[\frac{\tan 69^\circ + \tan 66^\circ}{1 - \tan 69^\circ \tan 66^\circ} = - 1\].
If \[\tan A = \frac{5}{6}\text{ and }\tan B = \frac{1}{11}\], prove that \[A + B = \frac{\pi}{4}\].
If \[\tan A = \frac{m}{m - 1}\text{ and }\tan B = \frac{1}{2m - 1}\], then prove that \[A - B = \frac{\pi}{4}\].
Prove that:
\[\cos^2 45^\circ - \sin^2 15^\circ = \frac{\sqrt{3}}{4}\]
Prove that:
sin2 (n + 1) A − sin2 nA = sin (2n + 1) A sin A.
Prove that:
\[\frac{\sin \left( A - B \right)}{\cos A \cos B} + \frac{\sin \left( B - C \right)}{\cos B \cos C} + \frac{\sin \left( C - A \right)}{\cos C \cos A} = 0\]
Prove that:
sin2 B = sin2 A + sin2 (A − B) − 2 sin A cos B sin (A − B)
Prove that:
\[\frac{\tan \left( A + B \right)}{\cot \left( A - B \right)} = \frac{\tan^2 A - \tan^2 B}{1 - \tan^2 A \tan^2 B}\]
If tan A = x tan B, prove that
\[\frac{\sin \left( A - B \right)}{\sin \left( A + B \right)} = \frac{x - 1}{x + 1}\]
If x lies in the first quadrant and \[\cos x = \frac{8}{17}\], then prove that:
Find the maximum and minimum values of each of the following trigonometrical expression:
12 cos x + 5 sin x + 4
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 tan \[\alpha = \frac{1}{1 + 2^{- x}}\] and \[\tan \beta = \frac{1}{1 + 2^{x + 1}}\] then write the value of α + β lying in the interval \[\left( 0, \frac{\pi}{2} \right)\]
If in ∆ABC, tan A + tan B + tan C = 6, then cot A cot B cot C =
If sin (π cos x) = cos (π sin x), then sin 2x = ______.
If tan (π/4 + x) + tan (π/4 − x) = a, then tan2 (π/4 + x) + tan2 (π/4 − x) =
If tan (A − B) = 1 and sec (A + B) = \[\frac{2}{\sqrt{3}}\], the smallest positive value of B is
If \[\tan\alpha = \frac{x}{x + 1}\] and \[\tan\alpha = \frac{x}{x + 1}\], then \[\alpha + \beta\] is equal to
Show that 2 sin2β + 4 cos (α + β) sin α sin β + cos 2(α + β) = cos 2α
Match each item given under column C1 to its correct answer given under column C2.
| C1 | C2 |
| (a) `(1 - cosx)/sinx` | (i) `cot^2 x/2` |
| (b) `(1 + cosx)/(1 - cosx)` | (ii) `cot x/2` |
| (c) `(1 + cosx)/sinx` | (iii) `|cos x + sin x|` |
| (d) `sqrt(1 + sin 2x)` | (iv) `tan x/2` |
If sin(θ + α) = a and sin(θ + β) = b, then prove that cos 2(α - β) - 4ab cos(α - β) = 1 - 2a2 - 2b2
[Hint: Express cos(α - β) = cos((θ + α) - (θ + β))]
If cos(θ + Φ) = m cos(θ – Φ), then prove that 1 tan θ = `(1 - m)/(1 + m) cot phi`
[Hint: Express `(cos(theta + Φ))/(cos(theta - Φ)) = m/1` and apply Componendo and Dividendo]
If tan θ = 3 and θ lies in third quadrant, then the value of sin θ ______.
If tanθ = `a/b`, then bcos2θ + asin2θ is equal to ______.
If sinx + cosx = a, then |sinx – cosx| = ______.
The maximum distance of a point on the graph of the function y = `sqrt(3)` sinx + cosx from x-axis is ______.
State whether the statement is True or False? Also give justification.
If tanA = `(1 - cos B)/sinB`, then tan2A = tanB
