Advertisements
Advertisements
Question
If \[T_n = \sin^n x + \cos^n x\], prove that \[\frac{T_3 - T_5}{T_1} = \frac{T_5 - T_7}{T_3}\]
Solution
LHS:\[\frac{T_3 - T_5}{T_1} = \frac{\left( \sin^3 x + \cos^3 x \right) - \left( \sin^5 x + \cos^5 x \right)}{\sin x + \cos x}\]
\[ = \frac{\sin^3 x - \sin^5 x + \cos^3 x - \cos^5 x}{\sin x + \cos x}\]
\[ = \frac{\sin^3 x\left( 1 - \sin^2 x \right) + \cos^3 x\left( 1 - \cos^2 x \right)}{\sin x + \cos x}\]
\[ = \frac{\sin^3 x . \cos^2 x + c {os}^3 x . \sin^2 x}{\sin x + \cos x}\]
\[ = \frac{\sin^2 x . \cos^2 x\left( \sin x + cos x \right)}{\sin x + \cos x}\]
\[ = \sin^2 x . \cos^2 x\]
RHS: \[\frac{T_5 - T_7}{T_3}\]
\[ = \frac{\left( \sin^5 x + \cos^5 x \right) - \left( \sin^7 x + \cos^7 x \right)}{\sin^3 x + \cos^3 x}\]
\[ = \frac{\sin^5 x - si n^7 x + \cos^5 x - \cos^7 x}{\sin^3 x + \cos^3 x}\]
\[ = \frac{\sin^5 x\left( 1 - \sin^2 x \right) + \cos^5 x\left( 1 - \cos^2 x \right)}{\sin^3 x + \cos^3 x}\]
\[ = \frac{\sin^5 x \cos^2 x + \cos^5 x \sin^2 x}{\sin^3 x + \cos^3 x}\]
\[ = \sin^2 x . \cos^2 x\]
LHS = RHS
Hence proved.
APPEARS IN
RELATED QUESTIONS
If \[\sin x = \frac{a^2 - b^2}{a^2 + b^2}\], then the values of tan x, sec x and cosec x
If \[\cot x \left( 1 + \sin x \right) = 4 m \text{ and }\cot x \left( 1 - \sin x \right) = 4 n,\] \[\left( m^2 + n^2 \right)^2 = mn\]
If \[a = \sec x - \tan x \text{ and }b = cosec x + \cot x\], then shown that \[ab + a - b + 1 = 0\]
Prove the:
\[ \sqrt{\frac{1 - \sin x}{1 + \sin x}} + \sqrt{\frac{1 + \sin x}{1 - \sin x}} = - \frac{2}{\cos x},\text{ where }\frac{\pi}{2} < x < \pi\]
If \[T_n = \sin^n x + \cos^n x\], prove that \[6 T_{10} - 15 T_8 + 10 T_6 - 1 = 0\]
Prove that:
Prove that
In a ∆A, B, C, D be the angles of a cyclic quadrilateral, taken in order, prove that cos(180° − A) + cos (180° + B) + cos (180° + C) − sin (90° + D) = 0
Find x from the following equations:
\[x \cot\left( \frac{\pi}{2} + \theta \right) + \tan\left( \frac{\pi}{2} + \theta \right)\sin \theta + cosec\left( \frac{\pi}{2} + \theta \right) = 0\]
If sec \[x = x + \frac{1}{4x}\], then sec x + tan x =
If tan \[x = - \frac{1}{\sqrt{5}}\] and θ lies in the IV quadrant, then the value of cos x is
If \[\frac{3\pi}{4} < \alpha < \pi, \text{ then }\sqrt{2\cot \alpha + \frac{1}{\sin^2 \alpha}}\] is equal to
sin6 A + cos6 A + 3 sin2 A cos2 A =
If \[cosec x + \cot x = \frac{11}{2}\], then tan x =
sin2 π/18 + sin2 π/9 + sin2 7π/18 + sin2 4π/9 =
If tan A + cot A = 4, then tan4 A + cot4 A is equal to
The value of \[\tan1^\circ \tan2^\circ \tan3^\circ . . . \tan89^\circ\] is
Which of the following is correct?
Find the general solution of the following equation:
Find the general solution of the following equation:
Find the general solution of the following equation:
Solve the following equation:
Solve the following equation:
Solve the following equation:
Solve the following equation:
`cosec x = 1 + cot x`
Solve the following equation:
4sinx cosx + 2 sin x + 2 cosx + 1 = 0
The general value of x satisfying the equation
\[\sqrt{3} \sin x + \cos x = \sqrt{3}\]
If \[4 \sin^2 x = 1\], then the values of x are
A value of x satisfying \[\cos x + \sqrt{3} \sin x = 2\] is
The number of values of x in [0, 2π] that satisfy the equation \[\sin^2 x - \cos x = \frac{1}{4}\]
The solution of the equation \[\cos^2 x + \sin x + 1 = 0\] lies in the interval
Solve the following equations:
sin 5x − sin x = cos 3
Solve the following equations:
`sin theta + sqrt(3) cos theta` = 1
Solve the following equations:
2cos 2x – 7 cos x + 3 = 0
Solve `sqrt(3)` cos θ + sin θ = `sqrt(2)`
If sin θ and cos θ are the roots of the equation ax2 – bx + c = 0, then a, b and c satisfy the relation ______.
Find the general solution of the equation 5cos2θ + 7sin2θ – 6 = 0
