मराठी
महाराष्ट्र राज्य शिक्षण मंडळएस.एस.सी (इंग्रजी माध्यम) इयत्ता १० वी

Prove that (cos(90^circ – A))/(sin A) = (sin(90^circ – A))/(cos A).

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प्रश्न

Prove that `(cos(90^circ - A))/(sin A) = (sin(90^circ - A))/(cos A)`.

सिद्धांत
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उत्तर

L.H.S. = `(cos(90^circ - A))/(sin A)`

= `(sin A)/(sin A)`

= 1

R.H.S. = `(sin(90^circ - A))/(cos A)`

= `(cos A)/(cos A)`

= 1

∴ L.H.S. = R.H.S.

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पाठ 6: Trigonometry - Q.1 (B)

संबंधित प्रश्‍न

Prove the following trigonometric identities.

`cos A/(1 - tan A) + sin A/(1 - cot A)  = sin A + cos A`


Prove the following trigonometric identities.

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


If 3 sin θ + 5 cos θ = 5, prove that 5 sin θ – 3 cos θ = ± 3.


Prove the following identities:

`(1 - sinA)/(1 + sinA) = (secA - tanA)^2`


`(1-cos^2theta) sec^2 theta = tan^2 theta`


`1 + (tan^2 θ)/((1 + sec θ)) = sec θ`


If `(cot theta ) = m and ( sec theta - cos theta) = n " prove that " (m^2 n)(2/3) - (mn^2)(2/3)=1`


If cosec θ = 2x and \[5\left( x^2 - \frac{1}{x^2} \right)\] \[2\left( x^2 - \frac{1}{x^2} \right)\] 


If cot θ + b cosec θ = p and b cot θ − a cosec θ = q, then p2 − q2 


Prove the following identity :

sinθcotθ + sinθcosecθ = 1 + cosθ  


Prove the following identity  :

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


Prove the following identity : 

`(cosecθ)/(tanθ + cotθ) = cosθ`


Prove the following identity : 

`sin^8θ - cos^8θ = (sin^2θ - cos^2θ)(1 - 2sin^2θcos^2θ)`


Prove that cot θ. tan (90° - θ) - sec (90° - θ). cosec θ + 1 = 0.


Prove the following identities.

`sqrt((1 + sin theta)/(1 - sin theta)) + sqrt((1 - sin theta)/(1 + sin theta))` = 2 sec θ


If a cos θ – b sin θ = c, then prove that (a sin θ + b cos θ) = `±  sqrt(a^2 + b^2 - c^2)`


tan (90 – θ) = ?


If 5 sec θ – 12 cosec θ = 0, then find values of sin θ, sec θ.


If cosec A – sin A = p and sec A – cos A = q, then prove that `(p^2q)^(2/3) + (pq^2)^(2/3) = 1`.


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`


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