Advertisements
Advertisements
Question
Prove the following:
`tan 4x = (4tan x(1 - tan^2 x))/(1 - 6tan^2 x + tan^4 x)`
Advertisements
Solution
`tan4 = tan 2 (2x) = (2tan2x)/(1 - tan^2 2x)`
= `(2tan 2x)/(1 - tan^2 (2x))`
= `((2tanx)/(1 - tan^2x))/(1 - (2tan^x)/(1 - tan^2x) `
= `(4tanx (1 - tan^2 x))/((1 - tan^2x)^2 - 4 tan^2 x)`
= `(4tanx ( 1 - tan^2 x))/(1 - 2 tan^2x+ tan^2 x - 4tan^2`
= `(4tanx (1 - tan^2x))/(1 + tan^4 x - 6 tan^2x)`
APPEARS IN
RELATED QUESTIONS
Prove the following: `(tan(pi/4 + x))/(tan(pi/4 - x)) = ((1+ tan x)/(1- tan x))^2`
Prove the following:
`cos ((3pi)/4 + x) - cos((3pi)/4 - x) = -sqrt2 sin x`
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`
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:
If \[\sin A = \frac{3}{5}, \cos B = - \frac{12}{13}\], where A and B both lie in second quadrant, find the value of sin (A + B).
If \[\cos A = - \frac{24}{25}\text{ and }\cos B = \frac{3}{5}\], where π < A < \[\frac{3\pi}{2}\text{ and }\frac{3\pi}{2}\]< B < 2π, find the following:
sin (A + B)
If \[\cos A = - \frac{24}{25}\text{ and }\cos B = \frac{3}{5}\], where π < A < \[\frac{3\pi}{2}\text{ and }\frac{3\pi}{2}\]< B < 2π, find the following:
cos (A + B)
Evaluate the following:
sin 36° cos 9° + cos 36° sin 9°
Prove that:
\[\frac{7\pi}{12} + \cos\frac{\pi}{12} = \sin\frac{5\pi}{12} - \sin\frac{\pi}{12}\]
Prove that:
Prove that:
\[\cos^2 45^\circ - \sin^2 15^\circ = \frac{\sqrt{3}}{4}\]
Prove that: \[\frac{\sin \left( A + B \right) + \sin \left( A - B \right)}{\cos \left( A + B \right) + \cos \left( A - B \right)} = \tan A\]
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:
cos2 A + cos2 B − 2 cos A cos B cos (A + B) = sin2 (A + B)
Prove that:
\[\frac{\tan \left( A + B \right)}{\cot \left( A - B \right)} = \frac{\tan^2 A - \tan^2 B}{1 - \tan^2 A \tan^2 B}\]
Prove that:
tan 8x − tan 6x − tan 2x = tan 8x tan 6x tan 2x
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:
If sin (α + β) = 1 and sin (α − β) \[= \frac{1}{2}\], where 0 ≤ α, \[\beta \leq \frac{\pi}{2}\], then find the values of tan (α + 2β) and tan (2α + β).
Find the maximum and minimum values of each of the following trigonometrical expression:
\[5 \cos x + 3 \sin \left( \frac{\pi}{6} - x \right) + 4\]
Write the maximum value of 12 sin x − 9 sin2 x.
If tan (A + B) = p and tan (A − B) = q, then write the value of tan 2B.
If a = b \[\cos \frac{2\pi}{3} = c \cos\frac{4\pi}{3}\] then write the value of ab + bc + ca.
tan 20° + tan 40° + \[\sqrt{3}\] tan 20° tan 40° is equal to
If A + B + C = π, then \[\frac{\tan A + \tan B + \tan C}{\tan A \tan B \tan C}\] is equal to
If \[\tan\theta = \frac{1}{2}\] and \[\tan\phi = \frac{1}{3}\], then the value of \[\tan\phi = \frac{1}{3}\] is
If `(sin(x + y))/(sin(x - y)) = (a + b)/(a - b)`, then show that `tanx/tany = a/b` [Hint: Use Componendo and Dividendo].
Find the most general value of θ satisfying the equation tan θ = –1 and cos θ = `1/sqrt(2)`.
The value of tan 75° - cot 75° is equal to ______.
If tanα = `1/7`, tanβ = `1/3`, then cos2α is equal to ______.
If tanθ = `a/b`, then bcos2θ + asin2θ is equal to ______.
Given x > 0, the values of f(x) = `-3cos sqrt(3 + x + x^2)` lie in the interval ______.
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 cosecx = 1 + cotx then x = 2nπ, 2nπ + `pi/2`
