Advertisements
Advertisements
प्रश्न
If \[5 \sin \alpha = 3 \sin \left( \alpha + 2 \beta \right) \neq 0\] , then \[\tan \left( \alpha + \beta \right)\] is equal to
पर्याय
\[2 \tan \beta\]
\[3 \tan \beta\]
\[4 \tan \beta\]
\[6 \tan \beta\]
Advertisements
उत्तर
\[4 \tan \beta\]
\[ \text{ We have } , \]
\[5 \sin \alpha = 3 \sin \left( \alpha + 2 \beta \right)\]
\[ \Rightarrow \frac{5}{3} = \frac{\sin \left( \alpha + 2 \beta \right)}{\sin \alpha}\]
\[ \Rightarrow \frac{5 - 3}{5 + 3} = \frac{\sin \left( \alpha + 2 \beta \right) - \sin \alpha}{\sin \left( \alpha + 2 \beta \right) + \sin \alpha} \left( \text{ Using componendo and dividendo } \right)\]
\[ \Rightarrow \frac{2}{8} = \frac{\sin \left( \alpha + 2 \beta \right) - \sin \alpha}{\sin \left( \alpha + 2 \beta \right) + \sin \alpha}\]
\[ \Rightarrow \frac{1}{4} = \frac{2\cos\frac{\alpha + 2 \beta + \alpha}{2}\sin\frac{\alpha + 2 \beta - \alpha}{2}}{2\sin\frac{\alpha + 2 \beta + \alpha}{2}\cos\frac{\alpha + 2 \beta - \alpha}{2}}\]
\[ \Rightarrow \frac{1}{4} = \frac{\cos\left( \alpha + \beta \right) \sin \beta}{\sin\left( \alpha + \beta \right) \cos \beta}\]
\[ \Rightarrow \frac{1}{4} = \cot \left( \alpha + \beta \right) \tan \beta\]
\[ \Rightarrow \frac{1}{4} = \frac{1}{\tan \left( \alpha + \beta \right)}\tan \beta\]
\[ \therefore \tan \left( \alpha + \beta \right) = 4 \tan \beta\]
APPEARS IN
संबंधित प्रश्न
Prove that: \[\frac{\sin x + \sin 2x}{1 + \cos x + \cos 2x} = \tan x\]
Prove that: \[\cos^2 \frac{\pi}{8} + \cos^2 \frac{3\pi}{8} + \cos^2 \frac{5\pi}{8} + \cos^2 \frac{7\pi}{8} = 2\]
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: \[\cos 4x = 1 - 8 \cos^2 x + 8 \cos^4 x\]
Show that: \[3 \left( \sin x - \cos x \right)^4 + 6 \left( \sin x + \cos \right)^2 + 4 \left( \sin^6 x + \cos^6 x \right) = 13\]
Prove that: \[\cos^6 A - \sin^6 A = \cos 2A\left( 1 - \frac{1}{4} \sin^2 2A \right)\]
Prove that: \[\cos^6 A - \sin^6 A = \cos 2A\left( 1 - \frac{1}{4} \sin^2 2A \right)\]
Prove that: \[\cos 4x - \cos 4\alpha = 8 \left( \cos x - \cos \alpha \right) \left( \cos x + \cos \alpha \right) \left( \cos x - \sin \alpha \right) \left( \cos x + \sin \alpha \right)\]
If \[\cos x = - \frac{3}{5}\] and x lies in the IIIrd quadrant, find the values of \[\cos\frac{x}{2}, \sin\frac{x}{2}, \sin 2x\] .
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: \[\cos \frac{\pi}{65} \cos \frac{2\pi}{65} \cos\frac{4\pi}{65} \cos\frac{8\pi}{65} \cos\frac{16\pi}{65} \cos\frac{32\pi}{65} = \frac{1}{64}\]
If \[2 \tan \alpha = 3 \tan \beta,\] prove that \[\tan \left( \alpha - \beta \right) = \frac{\sin 2\beta}{5 - \cos 2\beta}\] .
If \[a \cos2x + b \sin2x = c\] has α and β as its roots, then prove that
(iii)\[\tan\left( \alpha + \beta \right) = \frac{b}{a}\]
Prove that: \[4 \left( \cos^3 10 °+ \sin^3 20° \right) = 3 \left( \cos 10°+ \sin 2° \right)\]
Prove that \[\left| \sin x \sin \left( \frac{\pi}{3} - x \right) \sin \left( \frac{\pi}{3} + x \right) \right| \leq \frac{1}{4}\] for all values of x
Prove that: \[\sin^2 \frac{2\pi}{5} - \sin^{2 -} \frac{\pi}{3} = \frac{\sqrt{5} - 1}{8}\]
Prove that: \[\sin^2 24°- \sin^2 6° = \frac{\sqrt{5} - 1}{8}\]
Prove that: \[\sin^2 42° - \cos^2 78 = \frac{\sqrt{5} + 1}{8}\]
Prove that: \[\cos 78° \cos 42° \cos 36° = \frac{1}{8}\]
If \[\tan\frac{x}{2} = \frac{m}{n}\] , then write the value of m sin x + n cos x.
Write the value of \[\cos^2 76° + \cos^2 16° - \cos 76° \cos 16°\]
If \[\frac{\pi}{4} < x < \frac{\pi}{2}\], then write the value of \[\sqrt{1 - \sin 2x}\] .
If \[\text{ tan } A = \frac{1 - \text{ cos } B}{\text{ sin } B}\]
, then find the value of tan2A.
The value of \[2 \tan \frac{\pi}{10} + 3 \sec \frac{\pi}{10} - 4 \cos \frac{\pi}{10}\] is
The value of \[\left( \cot \frac{x}{2} - \tan \frac{x}{2} \right)^2 \left( 1 - 2 \tan x \cot 2 x \right)\] is
\[\sin^2 \left( \frac{\pi}{18} \right) + \sin^2 \left( \frac{\pi}{9} \right) + \sin^2 \left( \frac{7\pi}{18} \right) + \sin^2 \left( \frac{4\pi}{9} \right) =\]
\[2 \text{ cos } x - \ cos 3x - \cos 5x - 16 \cos^3 x \sin^2 x\]
The value of \[\frac{\cos 3x}{2 \cos 2x - 1}\] is equal to
If \[\tan \frac{x}{2} = \frac{\sqrt{1 - e}}{1 + e} \tan \frac{\alpha}{2}\] , then \[\cos \alpha =\]
If A = cos2θ + sin4θ for all values of θ, then prove that `3/4` ≤ A ≤ 1.
The greatest value of sin x cos x is ______.
The value of sin 20° sin 40° sin 60° sin 80° is ______.
Prove that sin 4A = 4sinA cos3A – 4 cosA sin3A
Find the value of the expression `cos^4 pi/8 + cos^4 (3pi)/8 + cos^4 (5pi)/8 + cos^4 (7pi)/8`
[Hint: Simplify the expression to `2(cos^4 pi/8 + cos^4 (3pi)/8) = 2[(cos^2 pi/8 + cos^2 (3pi)/8)^2 - 2cos^2 pi/8 cos^2 (3pi)/8]`
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 ______.
