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प्रश्न
If sinθ = `8/17`, where θ is an acute angle, find the value of cos θ by using identities.
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उत्तर
Given: sinθ = `8/17`
We know that sin2θ + cos2θ = 1
∴ cos2θ = 1 - sin2θ
∴ cos θ = `sqrt(1 - sin^2θ)`
Using given
`cosθ = sqrt(1 - (8/17)^2) = sqrt(1 - 8^2/17^2)`
∴ cos θ = `sqrt((17^2 - 8^2)/17^2) = sqrt(289 - 64)/17`
∴ cos θ = `sqrt(225)/17 = 15/17`
∴ cos θ = `15/17`
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संबंधित प्रश्न
If \[\sin\theta = \frac{7}{25}\], find the values of cosθ and tanθ.
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Prove that:
If \[\tan\theta + \frac{1}{\tan\theta} = 2\], then show that \[\tan^2 \theta + \frac{1}{\tan^2 \theta} = 2\]
Choose the correct alternative answer for the following question.
sin \[\theta\] cosec \[\theta\]= ?
Choose the correct alternative answer for the following question.
Prove the following.
secθ (1 – sinθ) (secθ + tanθ) = 1
Prove the following.
(secθ + tanθ) (1 – sinθ) = cosθ
Prove the following.
sec2θ + cosec2θ = sec2θ × cosec2θ
Prove the following.
\[\frac{\tan\theta}{\sec\theta + 1} = \frac{\sec\theta - 1}{\tan\theta}\]
Choose the correct alternative:
sinθ × cosecθ =?
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Prove that: (sec θ – cos θ) (cot θ + tan θ) = tan θ.sec θ
Proof: L.H.S. = (sec θ – cos θ) (cot θ + tan θ)
= `(1/square - cos θ) (square/square + square/square)` ......`[∵ sec θ = 1/square, cot θ = square/square and tan θ = square/square]`
= `((1 - square)/square) ((square + square)/(square square))`
= `square/square xx 1/(square square)` ......`[(∵ square + square = 1),(∴ square = 1 - square)]`
= `square/(square square)`
= tan θ.sec θ
= R.H.S.
∴ L.H.S. = R.H.S.
∴ (sec θ – cos θ) (cot θ + tan θ) = tan θ.sec θ
