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Complete the following activity to prove: cotθ + tanθ = cosecθ × secθ Activity: L.H.S. = cotθ + tanθ = θθθcosθsinθ+□cosθ = θθ□+sin2θsinθ×cosθ = θθ1sinθ× cosθ ....... ∵ □ = θθ1sinθ×1cosθ = θ□×secθ - Geometry Mathematics 2

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प्रश्न

Complete the following activity to prove:

cotθ + tanθ = cosecθ × secθ

Activity: L.H.S. = cotθ + tanθ

= `cosθ/sinθ + square/cosθ`

= `(square + sin^2theta)/(sinθ xx cosθ)`

= `1/(sinθ xx  cosθ)` ....... ∵ `square`

= `1/sinθ xx 1/cosθ`

= `square xx secθ`

∴ L.H.S. = R.H.S.

रिक्त स्थान भरें
योग

उत्तर

L.H.S. = cotθ + tanθ

= `cosθ/sinθ + sinθ/cosθ`

= `(cos^2θ + sin^2theta)/(sinθ xx cosθ)`

= `1/(sinθ xx  cosθ)` ....... ∵ sin2θ cos2θ = 1

= `1/sinθ xx 1/cosθ`

= cosecθ × secθ

∴ L.H.S. = R.H.S.

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2021-2022 (March) Set 1

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संबंधित प्रश्न

Prove the following trigonometric identities.

`1/(1 + sin A) + 1/(1 - sin A) =  2sec^2 A`


Prove the following trigonometric identities.

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


Prove the following trigonometric identities.

sin2 A cos2 B − cos2 A sin2 B = sin2 A − sin2 B


Prove the following identities:

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


Prove the following identities:

`tan^2A - tan^2B = (sin^2A - sin^2B)/(cos^2A * cos^2B)`


Prove the following identities:

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


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


`{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)`


Write the value of `(1 + cot^2 theta ) sin^2 theta`. 


Write the value of `(1+ tan^2 theta ) ( 1+ sin theta ) ( 1- sin theta)`


If 3 `cot theta = 4 , "write the value of" ((2 cos theta - sin theta))/(( 4 cos theta - sin theta))`


Write the value of cos1° cos 2°........cos180° .


If sinθ = `11/61`, find the values of cosθ using trigonometric identity.


Prove the following identity : 

`(1 + cotA + tanA)(sinA - cosA) = secA/(cosec^2A) - (cosecA)/sec^2A`


If x = asecθ + btanθ and y = atanθ + bsecθ , prove that `x^2 - y^2 = a^2 - b^2`


Without using trigonometric table , evaluate : 

`cos90^circ + sin30^circ tan45^circ cos^2 45^circ`


Prove that `((tan 20°)/(cosec 70°))^2 + ((cot 20°)/(sec 70°))^2  = 1`


To prove cot θ + tan θ = cosec θ × sec θ, complete the activity given below.

Activity:

L.H.S = `square`

= `square/sintheta + sintheta/costheta`

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

= `1/(sintheta*costheta)`     ......`[cos^2theta + sin^2theta = square]`

= `1/sintheta xx 1/square`

= `square`

= R.H.S


Prove that `(tan(90 - theta) + cot(90 - theta))/("cosec"  theta)` = sec θ


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