मराठी

Prove That: Tan 4 π − Cos 3 π 2 − Sin 5 π 6 Cos 2 π 3 = 1 4 - Mathematics

Advertisements
Advertisements

प्रश्न

Prove that:
\[\tan 4\pi - \cos\frac{3\pi}{2} - \sin\frac{5\pi}{6}\cos\frac{2\pi}{3} = \frac{1}{4}\]

Advertisements

उत्तर

\[ 4\pi = 720^\circ, \frac{3\pi}{2} = 270^\circ, \frac{5\pi}{6} = 150^\circ, \frac{2\pi}{3} = 120^\circ\]
LHS = \[\tan\left( 720^\circ \right) - \cos\left( 270^\circ \right) - \sin\left( 150^\circ \right) \cos\left( 120^\circ \right)\]
\[ = \tan\left( 90^\circ \times 8 + 0^\circ \right) - \cos\left( 90^\circ \times 3 + 0^\circ \right) - \sin\left( 90^\circ \times 1 + 60^\circ \right) \cos\left( 90^\circ \times 1 + 30^\circ \right)\]
\[ = \tan\left( 0^\circ \right) - \sin\left( 0^\circ \right) - \cos\left( 60^\circ \right) \left[ - \sin\left( 30^\circ \right) \right]\]
\[ = \tan\left( 0^\circ \right) - \sin\left( 0^\circ \right) + \cos\left( 60^\circ \right) \sin\left( 30^\circ \right)\]
\[ = 0 - 0 + \frac{1}{2} \times \frac{1}{2}\]
\[ = \frac{1}{4}\]
 = RHS
Hence proved.

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 5: Trigonometric Functions - Exercise 5.3 [पृष्ठ ४०]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 11
पाठ 5 Trigonometric Functions
Exercise 5.3 | Q 9.1 | पृष्ठ ४०

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

Find the general solution for each of the following equations sec2 2x = 1– tan 2x


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


If \[cosec x - \sin x = a^3 , \sec x - \cos x = b^3\], then prove that \[a^2 b^2 \left( a^2 + b^2 \right) = 1\]


Prove that:
\[\sin \frac{13\pi}{3}\sin\frac{2\pi}{3} + \cos\frac{4\pi}{3}\sin\frac{13\pi}{6} = \frac{1}{2}\]


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

 


If \[0 < x < \frac{\pi}{2}\], and if \[\frac{y + 1}{1 - y} = \sqrt{\frac{1 + \sin x}{1 - \sin x}}\], then y is equal to


If \[cosec x + \cot x = \frac{11}{2}\], then tan x =

 


If A lies in second quadrant 3tan A + 4 = 0, then the value of 2cot A − 5cosA + sin A is equal to


If tan θ + sec θ =ex, then cos θ equals


If sec x + tan x = k, cos x =


If \[f\left( x \right) = \cos^2 x + \sec^2 x\], then


The value of \[\cos1^\circ \cos2^\circ \cos3^\circ . . . \cos179^\circ\] is

 

The value of \[\tan1^\circ \tan2^\circ \tan3^\circ . . . \tan89^\circ\] is

 

Find the general solution of the following equation:

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

Find the general solution of the following equation:

\[\tan x = - \frac{1}{\sqrt{3}}\]

Find the general solution of the following equation:

\[\cos 3x = \frac{1}{2}\]

Find the general solution of the following equation:

\[\sin 2x = \cos 3x\]

Find the general solution of the following equation:

\[\tan 2x \tan x = 1\]

Find the general solution of the following equation:

\[\tan mx + \cot nx = 0\]

Find the general solution of the following equation:

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

Solve the following equation:

\[\cos 4 x = \cos 2 x\]

Solve the following equation:

\[\sin x + \sin 5x = \sin 3x\]

Solve the following equation:

\[\cos x \cos 2x \cos 3x = \frac{1}{4}\]

Solve the following equation:

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

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:
\[2^{\sin^2 x} + 2^{\cos^2 x} = 2\sqrt{2}\]


Write the solution set of the equation 

\[\left( 2 \cos x + 1 \right) \left( 4 \cos x + 5 \right) = 0\] in the interval [0, 2π].

If \[4 \sin^2 x = 1\], then the values of x are

 


If \[\cot x - \tan x = \sec x\], then, x is equal to

 


If \[e^{\sin x} - e^{- \sin x} - 4 = 0\], then x =


Choose the correct alternative:
If tan 40° = λ, then `(tan 140^circ - tan 130^circ)/(1 + tan 140^circ *  tan 130^circ)` =


Solve the equation sin θ + sin 3θ + sin 5θ = 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 ______.


The minimum value of 3cosx + 4sinx + 8 is ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×