Advertisements
Advertisements
Question
If \[2 \tan \alpha = 3 \tan \beta, \text{ then } \tan \left( \alpha - \beta \right) =\]
Options
\[\frac{\sin 2 \beta}{5 - \cos 2 \beta}\]
- \[\frac{\cos 2 \beta}{5 - \cos 2 \beta}\]
\[\frac{\sin 2 \beta}{5 + \cos 2 \beta}\]
none of these
Advertisements
Solution
\[\frac{\sin 2 \beta}{5 - \cos 2 \beta}\]
\[\text{ Given } : \]
\[2\tan\alpha = 3\tan\beta\]
\[\text{ Now } , \]
\[\tan\left( \alpha - \beta \right) = \frac{tan\alpha - tan\beta}{1 + \text{ tan } \alpha \text{ tan } \beta}\]
\[ = \frac{\frac{3}{2}tan\beta - tan\beta}{1 + \left( \frac{3}{2}tan\beta \right)tan\beta}\]
\[ = \frac{3tan\beta - 2tan\beta}{2 + 3 \tan^2 \beta}\]
\[ = \frac{tan\beta}{2 + 3 \tan^2 \beta}\]
\[ = \frac{\frac{sin\beta}{cos\beta}}{2 + 3\frac{\sin^2 \beta}{\cos^2 \beta}}\]
\[ = \frac{sin\beta cos\beta}{2 \cos^2 \beta + 3 \sin^2 \beta}\]
\[ = \frac{sin\beta cos\beta}{2 + \sin^2 \beta}\]
\[ = \frac{2sin\beta cos\beta}{4 + 2 \sin^2 \beta}\]
\[ = \frac{\sin2\beta}{4 + 1 - \cos2\beta}\]
\[ \therefore \tan\left( \alpha - \beta \right) = \frac{\sin2\beta}{5 - \cos2\beta}\]
APPEARS IN
RELATED QUESTIONS
Prove that: \[\frac{\cos 2 x}{1 + \sin 2 x} = \tan \left( \frac{\pi}{4} - x \right)\]
Prove that: \[\frac{\cos x}{1 - \sin x} = \tan \left( \frac{\pi}{4} + \frac{x}{2} \right)\]
Prove that: \[\sin^2 \frac{\pi}{8} + \sin^2 \frac{3\pi}{8} + \sin^2 \frac{5\pi}{8} + \sin^2 \frac{7\pi}{8} = 2\]
Prove that: \[\cos^6 A - \sin^6 A = \cos 2A\left( 1 - \frac{1}{4} \sin^2 2A \right)\]
Prove that: \[\cot^2 x - \tan^2 x = 4 \cot 2 x \text{ cosec } 2 x\]
Prove that: \[\cot \frac{\pi}{8} = \sqrt{2} + 1\]
If \[\cos x = - \frac{3}{5}\] and x lies in IInd quadrant, find the values of sin 2x and \[\sin\frac{x}{2}\] .
Prove that: \[\cos\frac{\pi}{5}\cos\frac{2\pi}{5}\cos\frac{4\pi}{5}\cos\frac{8\pi}{5} = \frac{- 1}{16}\]
If \[\cos x = \frac{\cos \alpha + \cos \beta}{1 + \cos \alpha \cos \beta}\] , prove that \[\tan\frac{x}{2} = \pm \tan\frac{\alpha}{2}\tan\frac{\beta}{2}\]
Prove that: \[4 \left( \cos^3 10 °+ \sin^3 20° \right) = 3 \left( \cos 10°+ \sin 2° \right)\]
Prove that `tan x + tan (π/3 + x) - tan(π/3 - x) = 3tan 3x`
\[\tan x + \tan\left( \frac{\pi}{3} + x \right) - \tan\left( \frac{\pi}{3} - x \right) = 3 \tan 3x\]
\[\cot x + \cot\left( \frac{\pi}{3} + x \right) + \cot\left( \frac{2\pi}{3} + x \right) = 3 \cot 3x\]
Prove that: \[\sin^2 24°- \sin^2 6° = \frac{\sqrt{5} - 1}{8}\]
Prove that: \[\cos 36° \cos 42° \cos 60° \cos 78° = \frac{1}{16}\]
Prove that: \[\sin\frac{\pi}{5}\sin\frac{2\pi}{5}\sin\frac{3\pi}{5}\sin\frac{4\pi}{5} = \frac{5}{16}\]
Write the value of \[\cos\frac{\pi}{7} \cos\frac{2\pi}{7} \cos\frac{4\pi}{7} .\]
If \[\text{ sin } x + \text{ cos } x = a\], then find the value of
If \[\cos 2x + 2 \cos x = 1\] then, \[\left( 2 - \cos^2 x \right) \sin^2 x\] is equal to
For all real values of x, \[\cot x - 2 \cot 2x\] is equal to
If \[\cos x = \frac{1}{2} \left( a + \frac{1}{a} \right),\] and \[\cos 3 x = \lambda \left( a^3 + \frac{1}{a^3} \right)\] then \[\lambda =\]
\[\sin^2 \left( \frac{\pi}{18} \right) + \sin^2 \left( \frac{\pi}{9} \right) + \sin^2 \left( \frac{7\pi}{18} \right) + \sin^2 \left( \frac{4\pi}{9} \right) =\]
If \[A = 2 \sin^2 x - \cos 2x\] , then A lies in the interval
The value of \[\frac{\cos 3x}{2 \cos 2x - 1}\] is equal to
\[2 \left( 1 - 2 \sin^2 7x \right) \sin 3x\] is equal to
If \[\tan \frac{x}{2} = \frac{\sqrt{1 - e}}{1 + e} \tan \frac{\alpha}{2}\] , then \[\cos \alpha =\]
The value of \[\frac{\sin 5 \alpha - \sin 3\alpha}{\cos 5 \alpha + 2 \cos 4\alpha + \cos 3\alpha} =\]
If \[\text{ tan } x = \frac{a}{b}\], then \[b \cos 2x + a \sin 2x\]
The value of sin 20° sin 40° sin 60° sin 80° is ______.
Prove that sin 4A = 4sinA cos3A – 4 cosA sin3A
If tanθ + sinθ = m and tanθ – sinθ = n, then prove that m2 – n2 = 4sinθ tanθ
[Hint: m + n = 2tanθ, m – n = 2sinθ, then use m2 – n2 = (m + n)(m – n)]
The value of cos12° + cos84° + cos156° + cos132° is ______.
The value of `sin pi/18 + sin pi/9 + sin (2pi)/9 + sin (5pi)/18` is given by ______.
