Advertisements
Advertisements
प्रश्न
Show that: \[3 \left( \sin x - \cos x \right)^4 + 6 \left( \sin x + \cos \right)^2 + 4 \left( \sin^6 x + \cos^6 x \right) = 13\]
Advertisements
उत्तर
\[LHS = 3 \left( \text{ sin } x - \text{ cos } x \right)^4 + 6 \left( \text{ sin } x + \text{ cos } x \right)^2 + 4\left( \sin^6 x + \cos^6 x \right)\]
`3{(sinx-cosx)^2}^2+6(sin^2x+cos^2x+2sinx.cosx)+4{(sin^2x)^3+(cos^2)^3}`
= `3(sin^2x+cos^2x-2sinxcosx)^2+6(1+2sinxcosx)+4{(sin^2x+cos^2x)((sin^2x)^2+(cos^2x)^2-sin^2x+cos^2x)}` ...`["a"^3+"b"^3=("a"+"b")("a"^2+"b"^2-"ab")]`
= `3(1-2sinxcosx)^2+6(1+2sinxcosx)+4(sin^4x+cos^4-sin^2xcos^2x)`
= `3(1+4sin^2xcos^2x-4sinxcosx)+6+12sinxcosx+4{(sin^2x+cos^2x)^2-2cos^2xsin^2x-sin^2xcos^2x}`
= `3+12sin^2cos^2x-12sinxcosx+6+12sinxcosx+4{(sin^2x+cos^2)^2-sin^2xcos^2x}`
= `9+12sin^2cos^2x+4-12sin^2cos^2x`
= 13 = R.H.S
Hence proved.
APPEARS IN
संबंधित प्रश्न
Prove that: \[\frac{\sin 2x}{1 - \cos 2x} = cot x\]
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: \[\frac{\cos x}{1 - \sin x} = \tan \left( \frac{\pi}{4} + \frac{x}{2} \right)\]
Prove that: \[\cos 4x = 1 - 8 \cos^2 x + 8 \cos^4 x\]
Prove that: \[\cot^2 x - \tan^2 x = 4 \cot 2 x \text{ cosec } 2 x\]
Prove that \[\sin 3x + \sin 2x - \sin x = 4 \sin x \cos\frac{x}{2} \cos\frac{3x}{2}\]
Prove that: \[\cot \frac{\pi}{8} = \sqrt{2} + 1\]
If \[\sin x = \frac{\sqrt{5}}{3}\] and x lies in IInd quadrant, find the values of \[\cos\frac{x}{2}, \sin\frac{x}{2} \text{ and } \tan \frac{x}{2}\] .
If \[\cos x = \frac{4}{5}\] and x is acute, find tan 2x
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 \[2 \tan\frac{\alpha}{2} = \tan\frac{\beta}{2}\] , prove that \[\cos \alpha = \frac{3 + 5 \cos \beta}{5 + 3 \cos \beta}\]
If \[\sin \alpha = \frac{4}{5} \text{ and } \cos \beta = \frac{5}{13}\] , prove that \[\cos\frac{\alpha - \beta}{2} = \frac{8}{\sqrt{65}}\]
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: \[\cos 78° \cos 42° \cos 36° = \frac{1}{8}\]
If \[\frac{\pi}{2} < x < \pi\], then write the value of \[\frac{\sqrt{1 - \cos 2x}}{1 + \cos 2x}\] .
If \[\pi < x < \frac{3\pi}{2}\], then write the value of \[\sqrt{\frac{1 - \cos 2x}{1 + \cos 2x}}\] .
Write the value of \[\cos^2 76° + \cos^2 16° - \cos 76° \cos 16°\]
The value of \[2 \tan \frac{\pi}{10} + 3 \sec \frac{\pi}{10} - 4 \cos \frac{\pi}{10}\] is
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 \[\lambda =\]
If \[2 \tan \alpha = 3 \tan \beta, \text{ then } \tan \left( \alpha - \beta \right) =\]
If \[\sin \alpha + \sin \beta = a \text{ and } \cos \alpha - \cos \beta = b \text{ then } \tan \frac{\alpha - \beta}{2} =\]
\[\sin^2 \left( \frac{\pi}{18} \right) + \sin^2 \left( \frac{\pi}{9} \right) + \sin^2 \left( \frac{7\pi}{18} \right) + \sin^2 \left( \frac{4\pi}{9} \right) =\]
\[2 \text{ cos } x - \ cos 3x - \cos 5x - 16 \cos^3 x \sin^2 x\]
The value of \[\frac{\cos 3x}{2 \cos 2x - 1}\] is equal to
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
The greatest value of sin x cos x is ______.
The value of sin 20° sin 40° sin 60° sin 80° is ______.
The value of `cos pi/5 cos (2pi)/5 cos (4pi)/5 cos (8pi)/5` is ______.
If tan(A + B) = p, tan(A – B) = q, then show that tan 2A = `(p + q)/(1 - pq)`
If acos2θ + bsin2θ = c has α and β as its roots, then prove that tanα + tanβ = `(2b)/(a + c)`.
`["Hint: Use the identities" cos2theta = (1 - tan^2theta)/(1 + tan^2theta) "and" sin2theta = (2tantheta)/(1 + tan^2theta)]`.
If sinθ = `(-4)/5` and θ lies in the third quadrant then the value of `cos theta/2` is ______.
The value of `sin pi/18 + sin pi/9 + sin (2pi)/9 + sin (5pi)/18` is given by ______.
If tanA = `(1 - cos "B")/sin"B"`, then tan2A = ______.
