हिंदी

If Sin2 θ Cos2 θ (1 + Tan2 θ) (1 + Cot2 θ) = λ, Then Find the Value of λ.

Advertisements
Advertisements

प्रश्न

If sin2 θ cos2 θ (1 + tan2 θ) (1 + cot2 θ) = λ, then find the value of λ. 

योग
Advertisements

उत्तर

Given: 

`sin ^2θ cos^2 θ(1+tan^2 θ)(1+cot ^2θ)=λ`

`⇒ sin^2θ cos^2 θ sec^2 θ cosec^2θ=λ`

⇒`(sin^2 θ cosec^2θ )xx (cos^2θ sec^2 θ)= λ` 

⇒ `(sin^2θ xx 1/sin^2θ )(cos^2 θxx1/cos^2θ)=λ`

\[\Rightarrow \lambda = 1 \times 1 = 1\]

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 11: Trigonometric Identities - Exercise 11.3 [पृष्ठ ५५]

APPEARS IN

आर.डी. शर्मा Mathematics [English] Class 10
अध्याय 11 Trigonometric Identities
Exercise 11.3 | Q 21 | पृष्ठ ५५

संबंधित प्रश्न

Prove the following identities:

`(i) 2 (sin^6 θ + cos^6 θ) –3(sin^4 θ + cos^4 θ) + 1 = 0`

`(ii) (sin^8 θ – cos^8 θ) = (sin^2 θ – cos^2 θ) (1 – 2sin^2 θ cos^2 θ)`


Prove the following identities, where the angles involved are acute angles for which the expressions are defined:

`(cos A-sinA+1)/(cosA+sinA-1)=cosecA+cotA ` using the identity cosec2 A = 1 cot2 A.


Prove the following trigonometric identities.

(1 + tan2θ) (1 − sinθ) (1 + sinθ) = 1


if `x/a cos theta + y/b sin theta = 1` and `x/a sin theta - y/b cos theta = 1` prove that `x^2/a^2 + y^2/b^2  = 2`


Prove the following identities:

`cosecA - cotA = sinA/(1 + cosA)`


`1/((1+tan^2 theta)) + 1/((1+ tan^2 theta))`


`1+ (cot^2 theta)/((1+ cosec theta))= cosec theta`


Prove the following identity :

`(1 - sin^2θ)sec^2θ = 1`


Prove the following identity :

`(cotA + tanB)/(cotB + tanA) = cotAtanB`


Prove the following identity :

`tanA - cotA = (1 - 2cos^2A)/(sinAcosA)`


Prove the following identity  :

`(1 + cotA)^2 + (1 - cotA)^2 = 2cosec^2A`


Prove that `(cos θ)/(1 - sin θ) = (1 + sin θ)/(cos θ)`.


Prove that sin4θ - cos4θ = sin2θ - cos2θ
= 2sin2θ - 1
= 1 - 2 cos2θ


Prove that: sin6θ + cos6θ = 1 - 3sin2θ cos2θ. 


Prove the following identities.

sec4 θ (1 – sin4 θ) – 2 tan2 θ = 1


1 + cot2θ = ? 


cos 45° = ?


Prove that cot2θ – tan2θ = cosec2θ – sec2θ.


If cot θ = `40/9`, find the values of cosec θ and sinθ,

We have, 1 + cot2θ = cosec2θ

1 + `square` = cosec2θ

1 + `square` = cosec2θ

`(square + square)/square` = cosec2θ

`square/square` = cosec2θ  ......[Taking root on the both side]

cosec θ = `41/9`

and sin θ = `1/("cosec"  θ)`

sin θ = `1/square`

∴ sin θ =  `9/41`

The value is cosec θ = `41/9`, and sin θ = `9/41`


sec θ when expressed in term of cot θ, is equal to ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×