मराठी

If \[F\Left( X \Right) = \Cos^2 X + \Sec^2 X\], Then

Advertisements
Advertisements

प्रश्न

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

पर्याय

  • f(x) < 1

  • f(x) = 1

  • 1 < f(x) < 2

  • f(x) ≥ 2

MCQ
Advertisements

उत्तर

\[f\left( x \right) = \cos^2 x + \sec^2 x\]
\[ = \cos^2 x + \sec^2 x - 2\cos x\sec x + 2\cos x\sec x\]
\[ = \left( \sec x - \cos x \right)^2 + 2\]
\[ \therefore f\left( x \right) \geq 2 \forall x \left[ \left( \sec x - \cos x \right)^2 \geq 0 \forall x \right]\]

Hence, the correct option is answer f(x) ≥ 2.

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 5: Trigonometric Functions - Exercise 5.5 [पृष्ठ ४३]

APPEARS IN

आर.डी. शर्मा Mathematics [English] Class 11
पाठ 5 Trigonometric Functions
Exercise 5.5 | Q 24 | पृष्ठ ४३

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

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


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


If \[\tan x = \frac{a}{b},\] show that

\[\frac{a \sin x - b \cos x}{a \sin x + b \cos x} = \frac{a^2 - b^2}{a^2 + b^2}\]

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


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

 


Prove that cos 570° sin 510° + sin (−330°) cos (−390°) = 0

Prove that

\[\frac{cosec(90^\circ + x) + \cot(450^\circ + x)}{cosec(90^\circ - x) + \tan(180^\circ - x)} + \frac{\tan(180^\circ + x) + \sec(180^\circ - x)}{\tan(360^\circ + x) - \sec( - x)} = 2\]

 


If sec \[x = x + \frac{1}{4x}\], then sec x + tan x = 

 

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


If tan \[x = - \frac{1}{\sqrt{5}}\] and θ lies in the IV quadrant, then the value of cos x is

 

If x sin 45° cos2 60° = \[\frac{\tan^2 60^\circ cosec30^\circ}{\sec45^\circ \cot^{2^\circ} 30^\circ}\], then x =

 

Find the general solution of the following equation:

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

Find the general solution of the following equation:

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

Solve the following equation:
\[\sin^2 x - \cos x = \frac{1}{4}\]


Solve the following equation:

\[4 \sin^2 x - 8 \cos x + 1 = 0\]

Solve the following equation:

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

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:

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


Write the general solutions of tan2 2x = 1.

 

Write the set of values of a for which the equation

\[\sqrt{3} \sin x - \cos x = a\] has no solution.

If \[3\tan\left( x - 15^\circ \right) = \tan\left( x + 15^\circ \right)\] \[0 < x < 90^\circ\], find θ.


If \[2 \sin^2 x = 3\cos x\]. where \[0 \leq x \leq 2\pi\], then find the value of x.


The general solution of the equation \[7 \cos^2 x + 3 \sin^2 x = 4\] is


The general value of x satisfying the equation
\[\sqrt{3} \sin x + \cos x = \sqrt{3}\]


The smallest positive angle which satisfies the equation ​

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

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


Solve the following equations for which solution lies in the interval 0° ≤ θ < 360°

2 sin2x + 1 = 3 sin x


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


Solve the following equations:
`sin theta + sqrt(3) cos theta` = 1


Choose the correct alternative:
If tan α and tan β are the roots of x2 + ax + b = 0 then `(sin(alpha + beta))/(sin alpha sin beta)` is equal to


Solve the equation sin θ + sin 3θ + sin 5θ = 0


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


If a cosθ + b sinθ = m and a sinθ - b cosθ = n, then show that a2 + b2 = m2 + n2 


Find the general solution of the equation 5cos2θ + 7sin2θ – 6 = 0


In a triangle ABC with ∠C = 90° the equation whose roots are tan A and tan B is ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×