हिंदी

If T N = Sin N X + Cos N X , Prove that T 3 − T 5 T 1 = T 5 − T 7 T 3

Advertisements
Advertisements

प्रश्न

If \[T_n = \sin^n x + \cos^n x\], prove that \[\frac{T_3 - T_5}{T_1} = \frac{T_5 - T_7}{T_3}\]

 

Advertisements

उत्तर

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.

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 5: Trigonometric Functions - Exercise 5.1 [पृष्ठ १९]

APPEARS IN

आर.डी. शर्मा Mathematics [English] Class 11
अध्याय 5 Trigonometric Functions
Exercise 5.1 | Q 26.1 | पृष्ठ १९

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

Find the principal and general solutions of the equation `tan x = sqrt3`


Find the principal and general solutions of the equation sec x = 2


Find the general solution of the equation sin 2x + cos x = 0


If \[\sin x = \frac{a^2 - b^2}{a^2 + b^2}\], then the values of tan x, sec x and cosec x


If \[a = \sec x - \tan x \text{ and }b = cosec x + \cot x\], then shown that  \[ab + a - b + 1 = 0\]


If \[T_n = \sin^n x + \cos^n x\], prove that \[6 T_{10} - 15 T_8 + 10 T_6 - 1 = 0\]


Prove that: \[\tan\frac{11\pi}{3} - 2\sin\frac{4\pi}{6} - \frac{3}{4} {cosec}^2 \frac{\pi}{4} + 4 \cos^2 \frac{17\pi}{6} = \frac{3 - 4\sqrt{3}}{2}\]

 


Prove that:
\[\frac{\cos (2\pi + x) cosec (2\pi + x) \tan (\pi/2 + x)}{\sec(\pi/2 + x)\cos x \cot(\pi + x)} = 1\]

 


Prove that

\[\left\{ 1 + \cot x - \sec\left( \frac{\pi}{2} + x \right) \right\}\left\{ 1 + \cot x + \sec\left( \frac{\pi}{2} + x \right) \right\} = 2\cot x\]

 


Prove that:
\[\sec\left( \frac{3\pi}{2} - x \right)\sec\left( x - \frac{5\pi}{2} \right) + \tan\left( \frac{5\pi}{2} + x \right)\tan\left( x - \frac{3\pi}{2} \right) = - 1 .\]


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 \[\frac{\pi}{2} < x < \frac{3\pi}{2},\text{ then }\sqrt{\frac{1 - \sin x}{1 + \sin x}}\] is equal to

 


If x = r sin θ cos ϕ, y = r sin θ sin ϕ and r cos θ, then x2 + y2 + z2 is independent of


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


If x is an acute angle and \[\tan x = \frac{1}{\sqrt{7}}\], then the value of \[\frac{{cosec}^2 x - \sec^2 x}{{cosec}^2 x + \sec^2 x}\] is


Find the general solution of the following equation:

\[\sin x = \frac{1}{2}\]

Find the general solution of the following equation:

\[\tan x + \cot 2x = 0\]

Solve the following equation:

\[2 \sin^2 x + \sqrt{3} \cos x + 1 = 0\]

Solve the following equation:

\[\cos x + \cos 3x - \cos 2x = 0\]

Solve the following equation:

\[\sin x + \sin 2x + \sin 3 = 0\]

Solve the following equation:

\[\sin x + \sin 2x + \sin 3x + \sin 4x = 0\]

Solve the following equation:

\[\sin 3x - \sin x = 4 \cos^2 x - 2\]

Solve the following equation:

\[\sin 2x - \sin 4x + \sin 6x = 0\]

Solve the following equation:

\[\tan x + \tan 2x = \tan 3x\]

Solve the following equation:
\[2 \sin^2 x = 3\cos x, 0 \leq x \leq 2\pi\]


Solve the following equation:
4sinx cosx + 2 sin x + 2 cosx + 1 = 0 


If secx cos5x + 1 = 0, where \[0 < x \leq \frac{\pi}{2}\], find the value of x.


If cos x = k has exactly one solution in [0, 2π], then write the values(s) of k.

 

A value of x satisfying \[\cos x + \sqrt{3} \sin x = 2\] is

 

The solution of the equation \[\cos^2 x + \sin x + 1 = 0\] lies in the interval


Find the principal solution and general solution of the following:
cot θ = `sqrt(3)`


Solve the following equations:
cos θ + cos 3θ = 2 cos 2θ


Solve the following equations:
sin θ + sin 3θ + sin 5θ = 0


Solve the following equations:
2cos 2x – 7 cos x + 3 = 0


Choose the correct alternative:
If f(θ) = |sin θ| + |cos θ| , θ ∈ R, then f(θ) is in the interval


If sin θ and cos θ are the roots of the equation ax2 – bx + c = 0, then a, b and c satisfy the relation ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×