Advertisements
Advertisements
Question
Prove that: \[\frac{\sin 2x}{1 - \cos 2x} = cot x\]
Advertisements
Solution
\[LHS = \frac{\sin2x}{1 - \cos2x}\]
\[ = \frac{2 sin x \times cos x}{2 \sin^2 x} \left( \because \sin2x , 1 - \cos2x = 2 \sin^2 x \right) \]
`= (2 sin x xx cos x)/(2 sin x xx sin x) `
`= (cos x)/(sin x) `
` = cot x = RHS`
\[\text{ Hence proved } .\]
APPEARS IN
RELATED QUESTIONS
Prove that: \[\frac{\cos 2 x}{1 + \sin 2 x} = \tan \left( \frac{\pi}{4} - x \right)\]
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: \[\left( \cos \alpha + \cos \beta^2 \right) + \left( \sin \alpha + \sin \beta \right)^2 = 4 \cos^2 \left( \frac{\alpha - \beta}{2} \right)\]
Prove that: \[1 + \cos^2 2x = 2 \left( \cos^4 x + \sin^4 x \right)\]
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 \[\sin 3x + \sin 2x - \sin x = 4 \sin x \cos\frac{x}{2} \cos\frac{3x}{2}\]
Prove that: \[\cos 7° \cos 14° \cos 28° \cos 56°= \frac{\sin 68°}{16 \cos 83°}\]
If \[\sin \alpha + \sin \beta = a \text{ and } \cos \alpha + \cos \beta = b\] , prove that
(ii) \[\cos \left( \alpha - \beta \right) = \frac{a^2 + b^2 - 2}{2}\]
If \[\sec \left( x + \alpha \right) + \sec \left( x - \alpha \right) = 2 \sec x\] , prove that \[\cos x = \pm \sqrt{2} \cos\frac{\alpha}{2}\]
If \[\cos \alpha + \cos \beta = \frac{1}{3}\] and sin \[\sin\alpha + \sin \beta = \frac{1}{4}\] , prove that \[\cos\frac{\alpha - \beta}{2} = \pm \frac{5}{24}\]
Prove that: \[\sin 5x = 5 \sin x - 20 \sin^3 x + 16 \sin^5 x\]
\[\cot x + \cot\left( \frac{\pi}{3} + x \right) + \cot\left( \frac{2\pi}{3} + x \right) = 3 \cot 3x\]
Prove that: \[\sin^2 24°- \sin^2 6° = \frac{\sqrt{5} - 1}{8}\]
Prove that: \[\sin^2 42° - \cos^2 78 = \frac{\sqrt{5} + 1}{8}\]
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: \[\cos\frac{\pi}{15} \cos \frac{2\pi}{15} \cos \frac{3\pi}{15} \cos \frac{4\pi}{15} \cos \frac{5\pi}{15} \cos\frac{6\pi}{15} \cos \frac{7\pi}{15} = \frac{1}{128}\]
If \[\cos 4x = 1 + k \sin^2 x \cos^2 x\] , then write the value of k.
If \[\frac{\pi}{2} < x < \pi,\] the write the value of \[\sqrt{2 + \sqrt{2 + 2 \cos 2x}}\] in the simplest form.
Write the value of \[\cos\frac{\pi}{7} \cos\frac{2\pi}{7} \cos\frac{4\pi}{7} .\]
If in a \[∆ ABC, \tan A + \tan B + \tan C = 0\], then
If \[\tan \alpha = \frac{1 - \cos \beta}{\sin \beta}\] , then
The value of \[\tan x \sin \left( \frac{\pi}{2} + x \right) \cos \left( \frac{\pi}{2} - x \right)\]
The value of \[2 \sin^2 B + 4 \cos \left( A + B \right) \sin A \sin B + \cos 2 \left( A + B \right)\] is
If \[\tan \frac{x}{2} = \frac{\sqrt{1 - e}}{1 + e} \tan \frac{\alpha}{2}\] , then \[\cos \alpha =\]
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 \[\tan x + \tan \left( \frac{\pi}{3} + x \right) + \tan \left( \frac{2\pi}{3} + x \right)\] is
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 ______.
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]`
The value of `(1 - tan^2 15^circ)/(1 + tan^2 15^circ)` is ______.
If sinθ = `(-4)/5` and θ lies in the third quadrant then the value of `cos theta/2` is ______.
