Advertisements
Advertisements
Question
If tanθ = 1 then, find the value of
`(sinθ + cosθ)/(secθ + cosecθ)`
Advertisements
Solution
tanθ = 1 ...(Given)
We know that, tan45° = 1
∴ θ = 45º
Now,
sinθ = sin 45º = `1/sqrt2`
cosθ = cos 45º = `1/sqrt2`
secθ = sec 45º = `sqrt2`
cosecθ = cosec 45º = `sqrt2`
∴ `(sinθ + cosθ)/(secθ + cosecθ)`
⇒ `(1/sqrt2 + 1/sqrt2)/(sqrt2 + sqrt2)`
⇒ `(2/sqrt2)/(2sqrt2)`
⇒ `cancel2/sqrt2 × 1/(cancel2sqrt2)`
⇒ `1/2`
∴ `(sinθ + cosθ)/(secθ + cosecθ) = 1/2`
APPEARS IN
RELATED QUESTIONS
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:
Prove that:
(secθ - cosθ)(cotθ + tanθ) = tanθ.secθ.
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:
Choose the correct alternative answer for the following question.
sin \[\theta\] cosec \[\theta\]= ?
Choose the correct alternative answer for the following question.
1 + tan2 \[\theta\] = ?
Choose the correct alternative answer for the following question.
Prove the following.
(secθ + tanθ) (1 – sinθ) = cosθ
Prove the following.
cot2θ – tan2θ = cosec2θ – sec2θ
Prove the following.
Prove the following:
sec6x – tan6x = 1 + 3sec2x × tan2x
Prove the following.
\[\frac{\tan\theta}{\sec\theta + 1} = \frac{\sec\theta - 1}{\tan\theta}\]
Prove the following.
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
