Advertisements
Advertisements
Question
\[\tan x + \tan\left( \frac{\pi}{3} + x \right) - \tan\left( \frac{\pi}{3} - x \right) = 3 \tan 3x\]
Advertisements
Solution
\[\frac{\pi}{3} = 60°\]
\[LHS = \text{ tan } x + \tan\left( 60° + x \right) - \tan\left( 60° - x \right)\]
\[ = \text{ tan } x + \left( \frac{\tan60° + \text { tan } x}{1 - \tan60° \text{ tan } x} \right) - \left( \frac{\tan60° - \text{ tan } x}{1 + \tan60° \text{ tan } x} \right)\]
\[ \left[ \tan\left( x + y \right) = \frac{\text{ tan } x + \text{ tan } y}{1 - \text{ tan } x \text{ tan } y} \text{ and } \tan\left( x - y \right) = \frac{\text{ tan } x - \text{ tan } y}{1 + \text{ tan } x \text{ tan } y} \right]\]
\[ = \text{ tanx } + \frac{\sqrt{3} + \text{ tan } x}{1 - \sqrt{3} \text{ tan } x} - \frac{\sqrt{3} - \text{ tan } x}{1 + \sqrt{3} \text{ tan } x}\]
\[ = \text{ tan } x + \frac{\sqrt{3} + 3\text{ tan } x + \text{ tan } x + \sqrt{3} \tan^2 x + \sqrt{3} + 3\text{ tan } x + \text{ tan } x - \sqrt{3} \tan^2 x}{\left( 1 - \sqrt{3} \text{ tan } x \right)\left( 1 + \sqrt{3} \text{ tan } x \right)}\]
\[ = \text{ tan } x + \frac{8\text{ tan } x}{1 - 3 \tan^2 x}\]
\[= \frac{\text{ tan } x - 3 \tan^3 x + 8\text{ tan } x}{1 - 3 \tan^2 x}\]
\[ = \frac{9\text{ tan } x - 3 \tan^3 x}{1 - 3 \tan^2 x}\]
\[ = 3\left( \frac{3\text{ tan } x - \tan^3 x}{1 - 3 \tan^2 x} \right) \left( \because \tan3\theta = \frac{3tan\theta - \tan^3 \theta}{1 - 3 \tan^2 \theta} \right) \]
\[ = 3\tan3x\]
\[ = RHS\]
\[\text{ Hence proved .} \]
APPEARS IN
RELATED QUESTIONS
Prove that: \[\frac{\sin 2x}{1 - \cos 2x} = cot x\]
Prove that: \[\sqrt{2 + \sqrt{2 + 2 \cos 4x}} = 2 \text{ cos } x\]
Prove that: \[\frac{\cos 2 x}{1 + \sin 2 x} = \tan \left( \frac{\pi}{4} - x \right)\]
Prove that: \[1 + \cos^2 2x = 2 \left( \cos^4 x + \sin^4 x \right)\]
Prove that: \[\cos^3 2x + 3 \cos 2x = 4\left( \cos^6 x - \sin^6 x \right)\]
Prove that: \[\left( \sin 3x + \sin x \right) \sin x + \left( \cos 3x - \cos x \right) \cos x = 0\]
Prove that: \[\cos 4x = 1 - 8 \cos^2 x + 8 \cos^4 x\]
Show that: \[2 \left( \sin^6 x + \cos^6 x \right) - 3 \left( \sin^4 x + \cos^4 x \right) + 1 = 0\]
Prove that: \[\cot \frac{\pi}{8} = \sqrt{2} + 1\]
If \[\sin x = \frac{\sqrt{5}}{3}\] and x lies in IInd quadrant, find the values of \[\cos\frac{x}{2}, \sin\frac{x}{2} \text{ and } \tan \frac{x}{2}\] .
Prove that: \[\cos \frac{\pi}{65} \cos \frac{2\pi}{65} \cos\frac{4\pi}{65} \cos\frac{8\pi}{65} \cos\frac{16\pi}{65} \cos\frac{32\pi}{65} = \frac{1}{64}\]
If \[2 \tan \alpha = 3 \tan \beta,\] prove that \[\tan \left( \alpha - \beta \right) = \frac{\sin 2\beta}{5 - \cos 2\beta}\] .
Prove that: \[4 \left( \cos^3 10 °+ \sin^3 20° \right) = 3 \left( \cos 10°+ \sin 2° \right)\]
Prove that: \[\sin^2 24°- \sin^2 6° = \frac{\sqrt{5} - 1}{8}\]
Prove that: \[\cos 6° \cos 42° \cos 66° \cos 78° = \frac{1}{16}\]
Prove that: \[\cos\frac{\pi}{15} \cos \frac{2\pi}{15} \cos \frac{3\pi}{15} \cos \frac{4\pi}{15} \cos \frac{5\pi}{15} \cos\frac{6\pi}{15} \cos \frac{7\pi}{15} = \frac{1}{128}\]
If \[\cos 4x = 1 + k \sin^2 x \cos^2 x\] , then write the value of k.
If \[\frac{\pi}{2} < x < \frac{3\pi}{2}\] , then write the value of \[\sqrt{\frac{1 + \cos 2x}{2}}\]
If \[\frac{\pi}{2} < x < \pi,\] the write the value of \[\sqrt{2 + \sqrt{2 + 2 \cos 2x}}\] in the simplest form.
If \[\frac{\pi}{4} < x < \frac{\pi}{2}\], then write the value of \[\sqrt{1 - \sin 2x}\] .
The value of \[\cos \frac{\pi}{65} \cos \frac{2\pi}{65} \cos \frac{4\pi}{65} \cos \frac{8\pi}{65} \cos \frac{16\pi}{65} \cos \frac{32\pi}{65}\] is
The value of \[2 \tan \frac{\pi}{10} + 3 \sec \frac{\pi}{10} - 4 \cos \frac{\pi}{10}\] is
If \[5 \sin \alpha = 3 \sin \left( \alpha + 2 \beta \right) \neq 0\] , then \[\tan \left( \alpha + \beta \right)\] is equal to
\[2 \text{ cos } x - \ cos 3x - \cos 5x - 16 \cos^3 x \sin^2 x\]
If \[A = 2 \sin^2 x - \cos 2x\] , then A lies in the interval
The value of \[\frac{2\left( \sin 2x + 2 \cos^2 x - 1 \right)}{\cos x - \sin x - \cos 3x + \sin 3x}\] is
\[2 \left( 1 - 2 \sin^2 7x \right) \sin 3x\] is equal to
If α and β are acute angles satisfying \[\cos 2 \alpha = \frac{3 \cos 2 \beta - 1}{3 - \cos 2 \beta}\] , then tan α =
If \[\left( 2^n + 1 \right) x = \pi,\] then \[2^n \cos x \cos 2x \cos 2^2 x . . . \cos 2^{n - 1} x = 1\]
The value of \[\cos^4 x + \sin^4 x - 6 \cos^2 x \sin^2 x\] is
The value of \[\tan x \tan \left( \frac{\pi}{3} - x \right) \tan \left( \frac{\pi}{3} + x \right)\] is
If \[\tan\alpha = \frac{1}{7}, \tan\beta = \frac{1}{3}\], then
\[\cos2\alpha\] is equal to
If A = cos2θ + sin4θ for all values of θ, then prove that `3/4` ≤ A ≤ 1.
If θ lies in the first quadrant and cosθ = `8/17`, then find the value of cos(30° + θ) + cos(45° – θ) + cos(120° – θ).
If sinθ = `(-4)/5` and θ lies in the third quadrant then the value of `cos theta/2` is ______.
If k = `sin(pi/18) sin((5pi)/18) sin((7pi)/18)`, then the numerical value of k is ______.
If tanA = `(1 - cos "B")/sin"B"`, then tan2A = ______.
