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If tan ⁡θ = 7/24, then to find value of cos θ complete the activity given below. Activity: sec^2θ = 1 + □ ...[Fundamental tri. identity] sec^2θ = 1 + □^2 sec^2θ = 1 + □/576 sec^2θ = □/576 sec θ = □

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प्रश्न

If `tan θ = 7/24`, then to find value of cos θ complete the activity given below.

Activity:

sec2θ = 1 + `square`   ...[Fundamental tri. identity]

sec2θ = 1 + `square^2`

sec2θ = 1 + `square/576`

sec2θ = `square/576`

sec θ = `square` 

cos θ = `square   ...`[cos theta = 1/sectheta]`

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उत्तर

\[\text{sec}^2θ = 1 + \boxed{\text{tan}^2θ}\]   ...[Fundamental tri. identity]

∴ \[\text{sec}^2θ = 1 + \boxed{\frac{7}{24}}^2\]

∴ \[\text{sec}^2θ = 1 + \frac{\boxed{49}}{576}\]

∴ sec2θ =`(576 + 49)/576`

∴ \[\text{sec}^2θ = \frac{\boxed{625}}{576}\]

∴ \[\text{sec} \phantom{.}θ = \boxed{\frac{25}{24}}\]

∴ \[\text{cos} \phantom{.}θ = \boxed{\frac{24}{25}}\]   ...`[cos theta = 1/sectheta]`

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अध्याय 6: Trigonometry - Q.3 (A)

संबंधित प्रश्न

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


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Activity:

L.H.S. = `square`

= `cos^2θ xx square`   ...`[1 + tan^2θ = square]`

= `(cos θ xx square)^2`

= 12

= 1

= R.H.S.


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Activity:

L.H.S. = `square`

= `square/(sinθ) + (sinθ)/(cosθ)`

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= `1/(sinθ) xx 1/square`

= `square`

= R.H.S.


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