Advertisements
Advertisements
प्रश्न
Prove the following.
\[\frac{\tan\theta}{\sec\theta + 1} = \frac{\sec\theta - 1}{\tan\theta}\]
Advertisements
उत्तर
\[\frac{\tan\theta}{\sec\theta + 1}\]
\[ = \frac{\tan\theta}{\sec\theta + 1} \times \frac{\sec\theta - 1}{\sec\theta - 1}\]
\[ = \frac{\tan\theta\left( \sec\theta - 1 \right)}{\sec^2 \theta - 1}\]
\[ = \frac{\tan\theta\left( \sec\theta - 1 \right)}{\tan^2 \theta} \] ...(1 + tan2θ = sec2θ)
\[ = \frac{\sec\theta - 1}{\tan\theta}\]
APPEARS IN
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.
If 5 secθ – 12 cosecθ = 0, find the values of secθ, cosθ, and sinθ.
If tanθ = 1 then, find the value of
`(sinθ + cosθ)/(secθ + cosecθ)`
Prove that:
Prove that: `1/"sec θ − tan θ" = "sec θ + tan θ"`
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.
cosec 45° =?
Choose the correct alternative answer for the following question.
Prove the following:
sec6x – tan6x = 1 + 3sec2x × tan2x
Choose the correct alternative:
sinθ × cosecθ =?
Show that:
`sqrt((1-cos"A")/(1+cos"A"))=cos"ecA - cotA"`
In ΔPQR, ∠P = 30°, ∠Q = 60°, ∠R = 90° and PQ = 12 cm, then find PR and QR.
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
