Advertisements
Advertisements
Question
Prove that `tan x + tan (π/3 + x) - tan(π/3 - x) = 3tan 3x`
Advertisements
Solution
`tan x + tan (π/3 + x) - tan(π/3 - x) = 3tan 3x`
LHS = `tan x + tan (π/3 + x) - tan(π/3 - x)`
`"LHS" = tan x + ((tan (π/3) + tan x)/(1 - tan x tan (π/3))) - ((tan (π/3) - tan x)/(1 + tan x tan (π/3)))`
We know that,
`tan (A + B) = ((tan A + tan B)/(1 - tan A tan B))` and
`tan (A _ B) = ((tan A - tan B)/(1 + tan A tan B))`
So,
`"LHS" = tan x + ((sqrt3 + tan x)/(1 - sqrt3 tan x)) - ((sqrt3 - tan x)/(1 + sqrt3 tan x))`
`"LHS" = tan x + (((1 + sqrt3 tan x)(sqrt3 + tan x) - (1 - sqrt3 tan x)(sqrt3 - tan x))/((1 - sqrt3tan x)(1 + sqrt3 tanx)))`
Simplify and cancel the similar terms of different sign in the above expression
we get,
`"LHS" = tan x + ((0 + 6tan x + 2tan x + 0)/(1 - 3tan^2x))`
`"LHS" = tan x + ((8tan x)/(1 - 3tan^2x))`
`"LHS" = (tan x (1 - 3tan^2x) + 8tan x)/(1 - 3tan^2x)`
`"LHS" = (tan x - 3tan^3x + 8tan x)/(1 - 3tan^2x)`
`"LHS" = (9tan x - 3tan^3x)/(1 - 3tan^2x)`
`"LHS" = 3((3tan x - tan^3x)/(1 - 3tan^2x))`
`"LHS" = 3 tan 3x ...{tan 3x = (3tanx - tan^3x)/(1 - 3tan^2x)`
RHS = 3 tan 3x
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: \[\left( \cos \alpha + \cos \beta^2 \right) + \left( \sin \alpha + \sin \beta \right)^2 = 4 \cos^2 \left( \frac{\alpha - \beta}{2} \right)\]
Prove that: \[\sin^2 \left( \frac{\pi}{8} + \frac{x}{2} \right) - \sin^2 \left( \frac{\pi}{8} - \frac{x}{2} \right) = \frac{1}{\sqrt{2}} \sin x\]
Prove that: \[\left( \sin 3x + \sin x \right) \sin x + \left( \cos 3x - \cos x \right) \cos x = 0\]
Prove that: \[\cos 4x = 1 - 8 \cos^2 x + 8 \cos^4 x\]
Prove that: \[\cos^6 A - \sin^6 A = \cos 2A\left( 1 - \frac{1}{4} \sin^2 2A \right)\]
Prove that:\[\tan\left( \frac{\pi}{4} + x \right) + \tan\left( \frac{\pi}{4} - x \right) = 2 \sec 2x\]
Prove that: \[\cos 7° \cos 14° \cos 28° \cos 56°= \frac{\sin 68°}{16 \cos 83°}\]
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 \[\tan \left( \alpha - \beta \right) = \frac{\sin 2\beta}{5 - \cos 2\beta}\] .
If \[\sin \alpha + \sin \beta = a \text{ and } \cos \alpha + \cos \beta = b\] , prove that
(i)\[\sin \left( \alpha + \beta \right) = \frac{2ab}{a^2 + b^2}\]
If \[2 \tan\frac{\alpha}{2} = \tan\frac{\beta}{2}\] , prove that \[\cos \alpha = \frac{3 + 5 \cos \beta}{5 + 3 \cos \beta}\]
If \[\cos \alpha + \cos \beta = \frac{1}{3}\] and sin \[\sin\alpha + \sin \beta = \frac{1}{4}\] , prove that \[\cos\frac{\alpha - \beta}{2} = \pm \frac{5}{24}\]
If \[a \cos2x + b \sin2x = c\] has α and β as its roots, then prove that
(iii)\[\tan\left( \alpha + \beta \right) = \frac{b}{a}\]
If \[\cos\alpha + \cos\beta = 0 = \sin\alpha + \sin\beta\] , then prove that \[\cos2\alpha + \cos2\beta = - 2\cos\left( \alpha + \beta \right)\] .
Prove that: \[\sin 5x = 5 \sin x - 20 \sin^3 x + 16 \sin^5 x\]
Prove that: \[\sin^2 \frac{2\pi}{5} - \sin^{2 -} \frac{\pi}{3} = \frac{\sqrt{5} - 1}{8}\]
Prove that: \[\sin^2 42° - \cos^2 78 = \frac{\sqrt{5} + 1}{8}\]
Prove that: \[\cos\frac{\pi}{15}\cos\frac{2\pi}{15}\cos\frac{4\pi}{15}\cos\frac{7\pi}{15} = \frac{1}{16}\]
Prove that: \[\sin\frac{\pi}{5}\sin\frac{2\pi}{5}\sin\frac{3\pi}{5}\sin\frac{4\pi}{5} = \frac{5}{16}\]
If \[\tan\frac{x}{2} = \frac{m}{n}\] , then write the value of m sin x + n cos x.
If \[\frac{\pi}{2} < x < \frac{3\pi}{2}\] , then write the value of \[\sqrt{\frac{1 + \cos 2x}{2}}\]
If \[\pi < x < \frac{3\pi}{2}\], then write the value of \[\sqrt{\frac{1 - \cos 2x}{1 + \cos 2x}}\] .
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
If \[2 \tan \alpha = 3 \tan \beta, \text{ then } \tan \left( \alpha - \beta \right) =\]
If \[5 \sin \alpha = 3 \sin \left( \alpha + 2 \beta \right) \neq 0\] , then \[\tan \left( \alpha + \beta \right)\] is equal to
The value of \[\frac{\cos 3x}{2 \cos 2x - 1}\] is equal to
If \[\tan \left( \pi/4 + x \right) + \tan \left( \pi/4 - x \right) = \lambda \sec 2x, \text{ then } \]
\[\frac{\sin 3x}{1 + 2 \cos 2x}\] is equal to
The value of `cos^2 48^@ - sin^2 12^@` is ______.
The value of `sin pi/18 + sin pi/9 + sin (2pi)/9 + sin (5pi)/18` is given by ______.
If A lies in the second quadrant and 3tanA + 4 = 0, then the value of 2cotA – 5cosA + sinA is equal to ______.
The value of `(sin 50^circ)/(sin 130^circ)` is ______.
