Advertisements
Advertisements
प्रश्न
If cosec2 θ (1 + cos θ) (1 − cos θ) = λ, then find the value of λ.
Advertisements
उत्तर
Given:
`cosec^2θ (1+cosθ)(1-cosθ)=λ`
⇒ `cosec^2θ (1+cosθ)(1-cosθ)=λ`
⇒ `cosec^2θ(1-cos^2θ)=λ`
⇒`cosec^θ sin^2θ=λ`
⇒`1/sin^2θxx sin^2θ=λ`
⇒` 1=λ`
⇒`λ=1`
Thus, the value of λ is 1.
APPEARS IN
संबंधित प्रश्न
Prove the following identities, where the angles involved are acute angles for which the expressions are defined:
`(1+ secA)/sec A = (sin^2A)/(1-cosA)`
[Hint : Simplify LHS and RHS separately.]
Prove that `(sin theta)/(1-cottheta) + (cos theta)/(1 - tan theta) = cos theta + sin theta`
Prove the identity (sin θ + cos θ)(tan θ + cot θ) = sec θ + cosec θ.
Prove the following trigonometric identities.
`tan theta + 1/tan theta = sec theta cosec theta`
Prove the following trigonometric identities.
`(cos theta)/(cosec theta + 1) + (cos theta)/(cosec theta - 1) = 2 tan theta`
Prove the following identities:
(cos A + sin A)2 + (cos A – sin A)2 = 2
Prove that:
(sec A − tan A)2 (1 + sin A) = (1 − sin A)
Prove that:
(1 + tan A . tan B)2 + (tan A – tan B)2 = sec2 A sec2 B
`1+(tan^2 theta)/((1+ sec theta))= sec theta`
`cos^2 theta /((1 tan theta))+ sin ^3 theta/((sin theta - cos theta))=(1+sin theta cos theta)`
If `( cosec theta + cot theta ) =m and ( cosec theta - cot theta ) = n, ` show that mn = 1.
If `(cosec theta - sin theta )= a^3 and (sec theta - cos theta ) = b^3 , " prove that " a^2 b^2 ( a^2+ b^2 ) =1`
Write the value of `sin theta cos ( 90° - theta )+ cos theta sin ( 90° - theta )`.
Write the value of sin A cos (90° − A) + cos A sin (90° − A).
If A = 30°, verify that `sin 2A = (2 tan A)/(1 + tan^2 A)`.
Prove that: `(1 + cot^2 θ/(1 + cosec θ)) = cosec θ`.
Prove the following identities.
(sin θ + sec θ)2 + (cos θ + cosec θ)2 = 1 + (sec θ + cosec θ)2
Choose the correct alternative:
sec2θ – tan2θ =?
If sec θ + tan θ = `sqrt(3)`, complete the activity to find the value of sec θ – tan θ
Activity:
`square` = 1 + tan2θ ......[Fundamental trigonometric identity]
`square` – tan2θ = 1
(sec θ + tan θ) . (sec θ – tan θ) = `square`
`sqrt(3)*(sectheta - tan theta)` = 1
(sec θ – tan θ) = `square`
Show that `(cos^2(45^circ + θ) + cos^2(45^circ - θ))/(tan(60^circ + θ) tan(30^circ - θ)) = 1`
