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Question
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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Solution
\[\text{sec}^2θ = 1 + \boxed{\text{tan}^2θ}\] ...[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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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)
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