English

What is the Value of (1 + Cot2 θ) Sin2 θ? - Mathematics

Advertisements
Advertisements

Question

What is the value of (1 + cot2 θ) sin2 θ?

Answer in Brief
Advertisements

Solution

We have, 

`(1+cot^2 θ)sin^2θ= cosec^2θxxsin^2θ` 

`= (1/sinθ)^2 xx sin^2θ` 

= `1/sin^2θxxsin^2θ` 

`=1`

shaalaa.com
  Is there an error in this question or solution?
Chapter 11: Trigonometric Identities - Exercise 11.3 [Page 55]

APPEARS IN

RD Sharma Mathematics [English] Class 10
Chapter 11 Trigonometric Identities
Exercise 11.3 | Q 3 | Page 55

RELATED QUESTIONS

 

Evaluate

`(sin ^2 63^@ + sin^2 27^@)/(cos^2 17^@+cos^2 73^@)`

 

if `cosec theta - sin theta = a^3`, `sec theta - cos theta = b^3` prove that `a^2 b^2 (a^2 + b^2) = 1`


Prove that:

`cot^2A/(cosecA - 1) - 1 = cosecA`


`sin theta / ((1+costheta))+((1+costheta))/sin theta=2cosectheta` 


`(1+ cos  theta - sin^2 theta )/(sin theta (1+ cos theta))= cot theta`


If `(x/a sin a - y/b cos theta) = 1 and (x/a cos theta + y/b sin theta ) =1, " prove that "(x^2/a^2 + y^2/b^2 ) =2`


`If sin theta = cos( theta - 45° ),where   theta   " is   acute, find the value of "theta` .


Prove that secθ + tanθ =`(costheta)/(1-sintheta)`.


(cosec θ − sin θ) (sec θ − cos θ) (tan θ + cot θ) is equal to


If cos  \[9\theta\] = sin \[\theta\] and  \[9\theta\]  < 900 , then the value of tan \[6 \theta\] is


Prove that: 
(cosec θ - sinθ )(secθ - cosθ ) ( tanθ +cot θ) =1


Prove the following identity :

tanA+cotA=secAcosecA 


Prove the following identity :

`(cosecA - sinA)(secA - cosA)(tanA + cotA) = 1`


Prove the following identity : 

`(cosecθ)/(tanθ + cotθ) = cosθ`


Prove the following identity :

`tan^2θ/(tan^2θ - 1) + (cosec^2θ)/(sec^2θ - cosec^2θ) = 1/(sin^2θ - cos^2θ)`


Find the value of `θ(0^circ < θ < 90^circ)` if : 

`tan35^circ cot(90^circ - θ) = 1`


Prove that: `(sin θ - 2sin^3 θ)/(2 cos^3 θ - cos θ) = tan θ`.


Prove that cot2θ × sec2θ = cot2θ + 1


If a sinθ + b cosθ = c, then prove that a cosθ – b sinθ = `sqrt(a^2 + b^2 - c^2)`.


(1 + sin A)(1 – sin A) is equal to ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×