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

If sec θ = 257, find the value of tan θ.

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प्रश्न

If sec θ = `25/7`, find the value of tan θ.

Solution:

1 + tan2 θ = sec2 θ

∴ 1 + tan2 θ = `(25/7)^square`

∴ tan2 θ = `625/49 - square`

= `(625 - 49)/49`

= `square/49`

∴ tan θ = `square/7` ........(by taking square roots)

बेरीज
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उत्तर

1 + tan2 θ = sec2 θ

∴ 1 + tan2 θ = `(25/7)^2`

∴ tan2 θ = `625/49 - 1`

= `(625 - 49)/49`

= `576/49`

∴ tan θ = `24/7` ........(by taking square roots)

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संबंधित प्रश्‍न

Prove the following trigonometric identities.

`(1 + sin θ)/cos θ+ cos θ/(1 + sin θ) = 2 sec θ`


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(sec A + tan A − 1) (sec A − tan A + 1) = 2 tan A


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`(tan A + tan B)/(cot A + cot B) = tan A tan B`


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Prove the following identities: sec2 θ + cosec2 θ = sec2 θ cosec2 θ.


Prove the following identities.

`costheta/(1 + sintheta)` = sec θ – tan θ


Prove that `(tan^2 theta - 1)/(tan^2 theta + 1)` = 1 – 2 cos2θ


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If cot θ = `40/9`, find the values of cosec θ and sinθ,

We have, 1 + cot2θ = cosec2θ

1 + `square` = cosec2θ

1 + `square` = cosec2θ

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∴ sin θ =  `9/41`

The value is cosec θ = `41/9`, and sin θ = `9/41`


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