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प्रश्न
Choose the correct alternative answer for the following question.
cosec 45° =?
पर्याय
\[\frac{1}{2}\]
\[\sqrt{2}\]
\[\frac{\sqrt{3}}{2}\]
\[\frac{2}{\sqrt{3}}\]
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उत्तर
Hence, the correct answer is \[\sqrt{2}\] .
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संबंधित प्रश्न
If \[\tan \theta = \frac{3}{4}\], find the values of secθ and cosθ
If \[\cot\theta = \frac{40}{9}\], find the values of cosecθ and sinθ.
Prove that:
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Prove that:
Prove that:
Choose the correct alternative answer for the following question.
sin \[\theta\] cosec \[\theta\]= ?
Choose the correct alternative answer for the following question.
1 + tan2 \[\theta\] = ?
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.
cot2θ – tan2θ = cosec2θ – sec2θ
Prove the following:
sec6x – tan6x = 1 + 3sec2x × tan2x
Prove the following.
\[\frac{\tan\theta}{\sec\theta + 1} = \frac{\sec\theta - 1}{\tan\theta}\]
Choose the correct alternative:
sinθ × cosecθ =?
If sinθ = `8/17`, where θ is an acute angle, find the value of cos θ by using identities.
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
