English

If A lies in the second quadrant and 3tanA + 4 = 0, then the value of 2cotA – 5cosA + sinA is equal to ______. - Mathematics

Advertisements
Advertisements

Question

If A lies in the second quadrant and 3tanA + 4 = 0, then the value of 2cotA – 5cosA + sinA is equal to ______.

Options

  • `(-53)/10`

  • `23/10`

  • `37/10`

  • `7/10`

MCQ
Fill in the Blanks

Solution

If A lies in the second quadrant and 3tanA + 4 = 0, then the value of 2cotA – 5cosA + sinA is equal to `23/10`.

Explanation:

Given that: 3tanA + 4 = 0

A lies in the second quadrant


tanA = `(-4)/3`

cosA = `(-3)/5`   .......[A lies in second quadrant]

And sinA = `4/5` and cotA = `(-3)/4`

∴ 2cotA – 5cosA + sinA = `2((-3)/4) -5((-3)/5) + 4/5`

= `(-3)/2 + 3 + 4/5`

= `(-15 + 30 + 8)/10`

= `23/10`

shaalaa.com
Values of Trigonometric Functions at Multiples and Submultiples of an Angle
  Is there an error in this question or solution?
Chapter 3: Trigonometric Functions - Exercise [Page 58]

APPEARS IN

NCERT Exemplar Mathematics [English] Class 11
Chapter 3 Trigonometric Functions
Exercise | Q 55 | Page 58

RELATED QUESTIONS

Prove that:  \[\sqrt{\frac{1 - \cos 2x}{1 + \cos 2x}} = \tan x\]


Prove that:  \[\frac{\sin 2x}{1 + \cos 2x} = \tan x\]

 

Prove that:  \[\frac{\cos x}{1 - \sin x} = \tan \left( \frac{\pi}{4} + \frac{x}{2} \right)\]


Prove that: \[\sin^2 \frac{\pi}{8} + \sin^2 \frac{3\pi}{8} + \sin^2 \frac{5\pi}{8} + \sin^2 \frac{7\pi}{8} = 2\]


Prove that: \[\cos^3 2x + 3 \cos 2x = 4\left( \cos^6 x - \sin^6 x \right)\]


Prove that: \[\cos^2 \left( \frac{\pi}{4} - x \right) - \sin^2 \left( \frac{\pi}{4} - x \right) = \sin 2x\]


Prove that:\[\tan\left( \frac{\pi}{4} + x \right) + \tan\left( \frac{\pi}{4} - x \right) = 2 \sec 2x\]

 

Prove that: \[\cot^2 x - \tan^2 x = 4 \cot 2 x  \text{ cosec }  2 x\]

 

\[\tan 82\frac{1° }{2} = \left( \sqrt{3} + \sqrt{2} \right) \left( \sqrt{2} + 1 \right) = \sqrt{2} + \sqrt{3} + \sqrt{4} + \sqrt{6}\]

 


 If  \[\cos x = - \frac{3}{5}\]  and x lies in IInd quadrant, find the values of sin 2x and \[\sin\frac{x}{2}\] .

 

 


Prove that:  \[\cos 7°  \cos 14° \cos 28° \cos 56°= \frac{\sin 68°}{16 \cos 83°}\]

 

If \[2 \tan\frac{\alpha}{2} = \tan\frac{\beta}{2}\] , prove that \[\cos \alpha = \frac{3 + 5 \cos \beta}{5 + 3 \cos \beta}\]

 

 


If  \[\cos\alpha + \cos\beta = 0 = \sin\alpha + \sin\beta\] , then prove that \[\cos2\alpha + \cos2\beta = - 2\cos\left( \alpha + \beta \right)\] .

 

Prove that:  \[\cos^3 x \sin 3x + \sin^3 x \cos 3x = \frac{3}{4} \sin 4x\]

 

\[\cot x + \cot\left( \frac{\pi}{3} + x \right) + \cot\left( \frac{\pi}{3} - x \right) = 3 \cot 3x\]

 


Prove that: \[\sin^2 24°- \sin^2 6° = \frac{\sqrt{5} - 1}{8}\]

  

Prove that: \[\cos\frac{\pi}{15}\cos\frac{2\pi}{15}\cos\frac{4\pi}{15}\cos\frac{7\pi}{15} = \frac{1}{16}\]

 

Prove that : \[\sin\frac{\pi}{5}\sin\frac{2\pi}{5}\sin\frac{3\pi}{5}\sin\frac{4\pi}{5} = \frac{5}{16}\]

 

If \[\cos 2x + 2 \cos x = 1\]  then, \[\left( 2 - \cos^2 x \right) \sin^2 x\]  is equal to 

 
 

The value of  \[2 \tan \frac{\pi}{10} + 3 \sec \frac{\pi}{10} - 4 \cos \frac{\pi}{10}\] is 

 

The value of \[\left( \cot \frac{x}{2} - \tan \frac{x}{2} \right)^2 \left( 1 - 2 \tan x \cot 2 x \right)\] is 

 

The value of \[\frac{\cos 3x}{2 \cos 2x - 1}\]  is equal to

   

If \[\tan \left( \pi/4 + x \right) + \tan \left( \pi/4 - x \right) = \lambda \sec 2x, \text{ then } \]


The value of  \[2 \sin^2 B + 4 \cos \left( A + B \right) \sin A \sin B + \cos 2 \left( A + B \right)\] is 


Prove that sin 4A = 4sinA cos3A – 4 cosA sin3A


If tanθ + sinθ = m and tanθ – sinθ = n, then prove that m2 – n2 = 4sinθ tanθ 
[Hint: m + n = 2tanθ, m – n = 2sinθ, then use m2 – n2 = (m + n)(m – n)]


If θ lies in the first quadrant and cosθ = `8/17`, then find the value of cos(30° + θ) + cos(45° – θ) + cos(120° – θ).


The value of `(1 - tan^2 15^circ)/(1 + tan^2 15^circ)` is ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×