हिंदी

`{1/((Sec^2 Theta- Cos^2 Theta))+ 1/((Cosec^2 Theta - Sin^2 Theta))} ( Sin^2 Theta Cos^2 Theta) = (1- Sin^2 Theta Cos ^2 Theta)/(2+ Sin^2 Theta Cos^2 Theta)`

Advertisements
Advertisements

प्रश्न

`{1/((sec^2 theta- cos^2 theta))+ 1/((cosec^2 theta - sin^2 theta))} ( sin^2 theta cos^2 theta) = (1- sin^2 theta cos ^2 theta)/(2+ sin^2 theta cos^2 theta)`

Advertisements

उत्तर

LHS = `{1/((sec^2 theta- cos^2 theta))+ 1/((cosec^2 theta - sin^2 theta))} ( sin^2 theta cos^2 theta) `

       =`{(cos^2 theta)/(1- cos^4 theta)+ (sin^2 theta)/(1- sin^4 theta)}(sin^2 theta cos ^2 theta)`

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

     =`[cot^2 theta/(1+ cos^2 theta) + tan^2 theta/(1+ sin^2 theta)]sin^2 theta cos^2 theta`

    =`(cos^4 theta)/(1+ cos^2 theta)+( sin^4 theta) / (1+ sin^2 theta)`

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

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

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

  =`(cos^4 theta (1+sin^2 theta )+ sin^4 theta (1+cos^2theta))/(1+ sin^2 theta + cos^2 theta + sin^2 theta cos ^2 theta )`

  =`(cos^4 theta cos^4 theta sin^2 theta+ sin^4 theta + sin^4 theta cos ^2 theta )/(1+1 sin^2 theta cos^2 theta)`

  =`(cos^4 theta + sin^4 theta + sin^2 theta cos^2 theta (sin^2 theta + cos^2 theta))/(2+ sin^2 theta cos^2 theta)`

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

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

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

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

    =RHS

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 13: Trigonometric identities - Exercises 1

APPEARS IN

आर.एस. अग्रवाल Mathematics [English] Class 10
अध्याय 13 Trigonometric identities
Exercises 1 | Q 33

संबंधित प्रश्न

Prove the following trigonometric identities.

(1 + tan2θ) (1 − sinθ) (1 + sinθ) = 1


Prove the following identities:

sec2 A . cosec2 A = tan2 A + cot2 A + 2


Prove the following identities:

`(sinAtanA)/(1 - cosA) = 1 + secA`


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


`(sin theta+1-cos theta)/(cos theta-1+sin theta) = (1+ sin theta)/(cos theta)`


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


Write the value of `cosec^2 theta (1+ cos theta ) (1- cos theta).`


 Write True' or False' and justify your answer  the following : 

The value of sin θ+cos θ is always greater than 1 .


cos4 A − sin4 A is equal to ______.


If x = acosθ , y = bcotθ , prove that `a^2/x^2 - b^2/y^2 = 1.`


prove that `1/(1 + cos(90^circ - A)) + 1/(1 - cos(90^circ - A)) = 2cosec^2(90^circ - A)`


If cosθ = `5/13`, then find sinθ. 


Prove that `(sin (90° - θ))/cos θ + (tan (90° - θ))/cot θ + (cosec (90° - θ))/sec θ = 3`.


If tan α = n tan β, sin α = m sin β, prove that cos2 α  = `(m^2 - 1)/(n^2 - 1)`.


Prove the following identities:
`1/(sin θ + cos θ) + 1/(sin θ - cos θ) = (2sin θ)/(1 - 2 cos^2 θ)`.


If cot θ + tan θ = x and sec θ – cos θ = y, then prove that `(x^2y)^(2/3) – (xy^2)^(2/3)` = 1


Prove that `cot^2 "A" [(sec "A" - 1)/(1 + sin "A")] + sec^2 "A" [(sin"A" - 1)/(1 + sec"A")]` = 0


Prove that `(cot A)/(1 - tan A) + (tan A)/(1 - cot A) = 1 + tan A + cot A = sec A  .  "cosec"  A + 1`.


If 5 tan β = 4, then `(5  sin β - 2 cos β)/(5 sin β + 2 cos β)` = ______.


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×