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प्रश्न
Choose the correct alternative:
sinθ × cosecθ =?
विकल्प
`sqrt(2)`
`1/2`
0
1
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उत्तर
sinθ × cosecθ = 1
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संबंधित प्रश्न
If \[\sin\theta = \frac{7}{25}\], find the values of cosθ and tanθ.
If \[\cot\theta = \frac{40}{9}\], find the values of cosecθ and sinθ.
If 5 secθ – 12 cosecθ = 0, find the values of secθ, cosθ, and sinθ.
Prove that:
(secθ - cosθ)(cotθ + tanθ) = tanθ.secθ.
Prove that: `1/"sec θ − tan θ" = "sec θ + tan θ"`
Prove that:
Prove that:
If \[\tan\theta + \frac{1}{\tan\theta} = 2\], then show that \[\tan^2 \theta + \frac{1}{\tan^2 \theta} = 2\]
Prove that:
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sin \[\theta\] cosec \[\theta\]= ?
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cosec 45° =?
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Prove the following.
secθ (1 – sinθ) (secθ + tanθ) = 1
Prove the following.
(secθ + tanθ) (1 – sinθ) = cosθ
Prove the following.
\[\frac{\tan\theta}{\sec\theta + 1} = \frac{\sec\theta - 1}{\tan\theta}\]
Prove the following.
Show that:
`sqrt((1-cos"A")/(1+cos"A"))=cos"ecA - cotA"`
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
