Advertisements
Advertisements
प्रश्न
For all real values of x, \[\cot x - 2 \cot 2x\] is equal to
विकल्प
- \[\tan 2x\]
- \[\tan x\]
- \[- \cot 3x\]
none of these
Advertisements
उत्तर
\[\text{ We have } , \]
\[\text{ cot } x - 2\cot 2x = \text{ cot } x - 2\frac{\cot^2 x - 1}{2\text{ cot } x}\]
\[ = \frac{\cot^2 x - \cot^2 x + 1}{\text{ cot } x}\]
\[ = \frac{1}{\text{ cot } x}\]
\[ = \text{ tan } x\]
APPEARS IN
संबंधित प्रश्न
Prove that: \[\frac{\sin 2x}{1 - \cos 2x} = cot x\]
Prove that: \[\frac{\sin 2x}{1 + \cos 2x} = \tan x\]
Prove that: \[\frac{1 - \cos 2x + \sin 2x}{1 + \cos 2x + \sin 2x} = \tan x\]
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: \[\left( \cos \alpha + \cos \beta^2 \right) + \left( \sin \alpha + \sin \beta \right)^2 = 4 \cos^2 \left( \frac{\alpha - \beta}{2} \right)\]
Prove that: \[\left( \sin 3x + \sin x \right) \sin x + \left( \cos 3x - \cos x \right) \cos x = 0\]
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: \[\cos 4x - \cos 4\alpha = 8 \left( \cos x - \cos \alpha \right) \left( \cos x + \cos \alpha \right) \left( \cos x - \sin \alpha \right) \left( \cos x + \sin \alpha \right)\]
If \[\sin x = \frac{4}{5}\] and \[0 < x < \frac{\pi}{2}\]
, find the value of sin 4x.
If \[\tan A = \frac{1}{7}\] and \[\tan B = \frac{1}{3}\] , show that cos 2A = sin 4B
Prove that: \[\cos 7° \cos 14° \cos 28° \cos 56°= \frac{\sin 68°}{16 \cos 83°}\]
Prove that: \[\cos\frac{2\pi}{15} \cos\frac{4\pi}{15} \cos \frac{8\pi}{15} \cos \frac{16\pi}{15} = \frac{1}{16}\]
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 \[2 \tan \alpha = 3 \tan \beta,\] prove that \[\tan \left( \alpha - \beta \right) = \frac{\sin 2\beta}{5 - \cos 2\beta}\] .
Prove that: \[\sin 5x = 5 \sin x - 20 \sin^3 x + 16 \sin^5 x\]
Prove that: \[\cos^3 x \sin 3x + \sin^3 x \cos 3x = \frac{3}{4} \sin 4x\]
Prove that `tan x + tan (π/3 + x) - tan(π/3 - x) = 3tan 3x`
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: \[\cos\frac{\pi}{15} \cos \frac{2\pi}{15} \cos \frac{3\pi}{15} \cos \frac{4\pi}{15} \cos \frac{5\pi}{15} \cos\frac{6\pi}{15} \cos \frac{7\pi}{15} = \frac{1}{128}\]
If \[\frac{\pi}{2} < x < \pi,\] the write the value of \[\sqrt{2 + \sqrt{2 + 2 \cos 2x}}\] in the simplest form.
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 \[\cos 2x + 2 \cos x = 1\] then, \[\left( 2 - \cos^2 x \right) \sin^2 x\] is equal to
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
If α and β are acute angles satisfying \[\cos 2 \alpha = \frac{3 \cos 2 \beta - 1}{3 - \cos 2 \beta}\] , then tan α =
The value of \[\tan x \tan \left( \frac{\pi}{3} - x \right) \tan \left( \frac{\pi}{3} + x \right)\] is
The value of \[\tan x + \tan \left( \frac{\pi}{3} + x \right) + \tan \left( \frac{2\pi}{3} + x \right)\] is
If \[\tan\alpha = \frac{1}{7}, \tan\beta = \frac{1}{3}\], then
\[\cos2\alpha\] is equal to
The greatest value of sin x cos x is ______.
Prove that sin 4A = 4sinA cos3A – 4 cosA sin3A
