Advertisements
Advertisements
प्रश्न
Prove that: \[\frac{\sin 2x}{1 + \cos 2x} = \tan x\]
Advertisements
उत्तर
\[LHS = \frac{\sin2x}{1 + \cos2x}\]
\[ = \frac{2\sin x \times \cos x}{1 + 2 \cos^2 x - 1} \] `[∵ sin 2 x = 2 sin x xx cos x and cos 2x = 2 cos ^2 x -1]`
`= (2 sin x xx cos x) /( 2 cos x xx cos x )`
` = (sin x) / (cos x) `
` = tan x = RHS`
\[\text{ Hence proved } .\]
APPEARS IN
संबंधित प्रश्न
Prove that: \[\frac{1 - \cos 2x + \sin 2x}{1 + \cos 2x + \sin 2x} = \tan x\]
Prove that: \[\frac{\sin x + \sin 2x}{1 + \cos x + \cos 2x} = \tan x\]
Prove that: \[\cos^2 \left( \frac{\pi}{4} - x \right) - \sin^2 \left( \frac{\pi}{4} - x \right) = \sin 2x\]
Prove that: \[\sin 4x = 4 \sin x \cos^3 x - 4 \cos x \sin^3 x\]
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:\[\tan\left( \frac{\pi}{4} + x \right) + \tan\left( \frac{\pi}{4} - x \right) = 2 \sec 2x\]
If 0 ≤ x ≤ π and x lies in the IInd quadrant such that \[\sin x = \frac{1}{4}\]. Find the values of \[\cos\frac{x}{2}, \sin\frac{x}{2} \text{ and } \tan\frac{x}{2}\]
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: \[\cos \frac{\pi}{65} \cos \frac{2\pi}{65} \cos\frac{4\pi}{65} \cos\frac{8\pi}{65} \cos\frac{16\pi}{65} \cos\frac{32\pi}{65} = \frac{1}{64}\]
If \[2 \tan \alpha = 3 \tan \beta,\] prove that \[\tan \left( \alpha - \beta \right) = \frac{\sin 2\beta}{5 - \cos 2\beta}\] .
If \[\sec \left( x + \alpha \right) + \sec \left( x - \alpha \right) = 2 \sec x\] , prove that \[\cos x = \pm \sqrt{2} \cos\frac{\alpha}{2}\]
If \[a \cos2x + b \sin2x = c\] has α and β as its roots, then prove that
(i) \[\tan\alpha + \tan\beta = \frac{2b}{a + c}\]
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: \[\sin 5x = 5 \sin x - 20 \sin^3 x + 16 \sin^5 x\]
Prove that `tan x + tan (π/3 + x) - tan(π/3 - x) = 3tan 3x`
\[\cot x + \cot\left( \frac{\pi}{3} + x \right) + \cot\left( \frac{2\pi}{3} + x \right) = 3 \cot 3x\]
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: \[\sin^2 \frac{2\pi}{5} - \sin^{2 -} \frac{\pi}{3} = \frac{\sqrt{5} - 1}{8}\]
Prove that: \[\sin^2 24°- \sin^2 6° = \frac{\sqrt{5} - 1}{8}\]
Prove that: \[\cos 78° \cos 42° \cos 36° = \frac{1}{8}\]
Prove that: \[\cos 6° \cos 42° \cos 66° \cos 78° = \frac{1}{16}\]
Prove that: \[\sin\frac{\pi}{5}\sin\frac{2\pi}{5}\sin\frac{3\pi}{5}\sin\frac{4\pi}{5} = \frac{5}{16}\]
If \[\frac{\pi}{2} < x < \frac{3\pi}{2}\] , then write the value of \[\sqrt{\frac{1 + \cos 2x}{2}}\]
If \[\frac{\pi}{2} < x < \pi\], then write the value of \[\frac{\sqrt{1 - \cos 2x}}{1 + \cos 2x}\] .
Write the value of \[\cos^2 76° + \cos^2 16° - \cos 76° \cos 16°\]
For all real values of x, \[\cot x - 2 \cot 2x\] is equal to
The value of \[\tan x \sin \left( \frac{\pi}{2} + x \right) \cos \left( \frac{\pi}{2} - x \right)\]
\[2 \left( 1 - 2 \sin^2 7x \right) \sin 3x\] is equal to
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 \[\tan x \tan \left( \frac{\pi}{3} - x \right) \tan \left( \frac{\pi}{3} + x \right)\] is
If \[n = 1, 2, 3, . . . , \text{ then } \cos \alpha \cos 2 \alpha \cos 4 \alpha . . . \cos 2^{n - 1} \alpha\] is equal to
If sinθ = `(-4)/5` and θ lies in the third quadrant then the value of `cos theta/2` is ______.
If A lies in the second quadrant and 3tanA + 4 = 0, then the value of 2cotA – 5cosA + sinA is equal to ______.
If k = `sin(pi/18) sin((5pi)/18) sin((7pi)/18)`, then the numerical value of k is ______.
