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Write the Value Of`(Tan^2 Theta - Sec^2 Theta)/(Cot^2 Theta - Cosec^2 Theta)` - Mathematics

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प्रश्न

Write the value of`(tan^2 theta  - sec^2 theta)/(cot^2 theta - cosec^2 theta)`

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उत्तर

`(tan^2 theta - sec^2 theta )/ (cot^2 theta - cosec^2 theta)`

  =` (-1)/(-1)`

  = 1

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अध्याय 8: Trigonometric Identities - Exercises 3

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आर.एस. अग्रवाल Mathematics [English] Class 10
अध्याय 8 Trigonometric Identities
Exercises 3 | Q 15

संबंधित प्रश्न

Prove the following trigonometric identities.

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


Prove the following trigonometric identities.

`[tan θ + 1/cos θ]^2 + [tan θ - 1/cos θ]^2 = 2((1 + sin^2 θ)/(1 - sin^2 θ))`


Prove that

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


If cos θ + cos2 θ = 1, prove that sin12 θ + 3 sin10 θ + 3 sin8 θ + sin6 θ + 2 sin4 θ + 2 sin2 θ − 2 = 1


Prove the following identities:

`((1 + tan^2A)cotA)/(cosec^2A) = tan A`


Prove the following identities:

`(1 - cosA)/sinA + sinA/(1 - cosA)= 2cosecA`


Prove the following identities:

cosec4 A (1 – cos4 A) – 2 cot2 A = 1


If `( sin theta + cos theta ) = sqrt(2) , " prove that " cot theta = ( sqrt(2)+1)`.


If \[sec\theta + tan\theta = x\] then \[tan\theta =\] 


Prove the following identity :

`(tanθ + sinθ)/(tanθ - sinθ) = (secθ + 1)/(secθ - 1)`


If `asin^2θ + bcos^2θ = c and p sin^2θ + qcos^2θ = r` , prove that (b - c)(r - p) = (c - a)(q - r)


Without using trigonometric identity , show that :

`sin(50^circ + θ) - cos(40^circ - θ) = 0`


Prove the following identities.

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


If `(cos alpha)/(cos beta)` = m and `(cos alpha)/(sin beta)` = n, then prove that (m2 + n2) cos2 β = n2


Prove that `sec"A"/(tan "A" + cot "A")` = sin A


Prove that sin4A – cos4A = 1 – 2cos2A


Prove that sec2θ – cos2θ = tan2θ + sin2θ


Prove that `sqrt(sec^2 theta + "cosec"^2 theta) = tan theta + cot theta`


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`


Eliminate θ if x = r cosθ and y = r sinθ.


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