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

If 3 sin θ = 4 cos θ, then sec θ = ?

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प्रश्न

If 3 sin θ = 4 cos θ, then sec θ = ?

बेरीज
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उत्तर

3 sin θ = 4 cos θ   ...[Given]

∴ `(sin θ)/(cos θ) = 4/3`

∴ `tan θ = 4/3`

We know that,

1 + tan2θ = sec2θ

∴  `1 + (4/3)^2 = sec^2θ`

∴ `1 + 16/9 = sec^2θ`

∴ `sec^2θ = (9 + 16)/9`

∴ `sec^2θ = 25/9`

∴ `sec θ = 5/3`   ...[Taking square root of both sides]

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पाठ 6: Trigonometry - Exercise

संबंधित प्रश्‍न

Express the ratios cos A, tan A and sec A in terms of sin A.


if `cos theta = 5/13` where `theta` is an acute angle. Find the value of `sin theta`


Prove that `(tan^2 theta)/(sec theta - 1)^2 = (1 + cos theta)/(1 - cos theta)`


Prove the following trigonometric identities.

tan2θ cos2θ = 1 − cos2θ


Prove that

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


Prove the following identities:

`1/(secA + tanA) = secA - tanA`


`sqrt((1+cos theta)/(1-cos theta)) + sqrt((1-cos theta )/(1+ cos theta )) = 2 cosec theta`

 


If `( cosec theta + cot theta ) =m and ( cosec theta - cot theta ) = n, ` show that mn = 1.


Write the value of `( 1- sin ^2 theta  ) sec^2 theta.`


Prove the following identity  :

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


Prove the following identity : 

`[1/((sec^2θ - cos^2θ)) + 1/((cosec^2θ - sin^2θ))](sin^2θcos^2θ) = (1 - sin^2θcos^2θ)/(2 + sin^2θcos^2θ)`


Prove that `(tan^2"A")/(tan^2 "A"-1) + (cosec^2"A")/(sec^2"A"-cosec^2"A") = (1)/(1-2 co^2 "A")`


Prove that sin2 θ + cos4 θ = cos2 θ + sin4 θ.


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


Prove that sin (90° - θ) cos (90° - θ) = tan θ. cos2θ.


Prove the following identities.

sec6 θ = tan6 θ + 3 tan2 θ sec2 θ + 1


If `sqrt(3)` sin θ – cos θ = θ, then show that tan 3θ = `(3tan theta - tan^3 theta)/(1 - 3 tan^2 theta)`


Show that tan4θ + tan2θ = sec4θ – 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`


(sec2 θ – 1) (cosec2 θ – 1) is equal to ______.


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