Advertisements
Advertisements
प्रश्न
If 5 sec θ – 12 cosec θ = 0, then find values of sin θ, sec θ.
Advertisements
उत्तर
5 sec θ – 12 cosec θ = 0 ...[Given]
∴ 5 sec θ = 12 cosec θ
∴ `5/(cosθ) = 12/(sinθ)` ...`[∵ sec θ = 1/(cosθ), "cosec" θ = 1/(sin θ)]`
∴ `(sinθ)/(cosθ) = 12/5`
∴ `tan θ = 12/5`
We know that,
1 + tan2θ = sec2θ
∴ `1 + (12/5)^2 = sec^2θ`
∴ `1 + 144/25 = sec^2θ`
∴ `(25 + 144)/25 = sec^2θ`
∴ `sec^2θ = 169/25`
∴ `sec θ = 13/5` ...[Taking square root of both sides]
Now, `cos θ = 1/(sec θ)`
= `1/((13/5))`
∴ `cos θ = 5/13`
We know that,
sin2θ + cos2θ = 1
∴ `sin^2θ + (5/13)^2 = 1`
∴ `sin^2θ + 25/169 = 1`
∴ `sec^2θ = 1 - 25/169`
∴ `sec^2θ = (169 - 25)/169`
∴ `sec^2θ = 144/169`
∴ `sin θ = 12/13` ...[Taking square root of both sides]
∴ `sin θ = 12/13, sec θ = 13/5`
APPEARS IN
संबंधित प्रश्न
If cosθ + sinθ = √2 cosθ, show that cosθ – sinθ = √2 sinθ.
Prove that (1 + cot θ – cosec θ)(1+ tan θ + sec θ) = 2
Prove the following trigonometric identities.
sin2 A cot2 A + cos2 A tan2 A = 1
Prove the following trigonometric identities.
`1/(sec A - 1) + 1/(sec A + 1) = 2 cosec A cot A`
Prove the following trigonometric identities.
`[tan θ + 1/cos θ]^2 + [tan θ - 1/cos θ]^2 = 2((1 + sin^2 θ)/(1 - sin^2 θ))`
Prove the following trigonometric identities.
(sec A + tan A − 1) (sec A − tan A + 1) = 2 tan A
Prove that `(sec theta - 1)/(sec theta + 1) = ((sin theta)/(1 + cos theta))^2`
`1/((1+tan^2 theta)) + 1/((1+ tan^2 theta))`
If m = ` ( cos theta - sin theta ) and n = ( cos theta + sin theta ) "then show that" sqrt(m/n) + sqrt(n/m) = 2/sqrt(1-tan^2 theta)`.
If tanθ `= 3/4` then find the value of secθ.
Write the value of \[\cot^2 \theta - \frac{1}{\sin^2 \theta}\]
(cosec θ − sin θ) (sec θ − cos θ) (tan θ + cot θ) is equal to
Prove the following Identities :
`(cosecA)/(cotA+tanA)=cosA`
Prove the following identity :
`(cos^3A + sin^3A)/(cosA + sinA) + (cos^3A - sin^3A)/(cosA - sinA) = 2`
Prove that:
`(sin A + cos A)/(sin A - cos A) + (sin A - cos A)/(sin A + cos A) = 2/(2 sin^2 A - 1)`
Prove that : `tan"A"/(1 - cot"A") + cot"A"/(1 - tan"A") = sec"A".cosec"A" + 1`.
sec 60° = ?
Prove that `(cos(90^circ - A))/(sin A) = (sin(90^circ - A))/(cos A)`.
If tan θ × A = sin θ, then A = ?
tan2θ – sin2θ = tan2θ × sin2θ. For proof of this complete the activity given below.
Activity:
L.H.S. = `square`
= `square (1 - (sin^2θ)/(tan^2θ))`
= `tan^2θ (1 - square/((sin^2θ)/(cos^2θ)))`
= `tan^2θ (1 - (sin^2θ)/1 xx (cos^2θ)/square)`
= `tan^2θ (1 - square)`
= `tan^2θ xx square` ...[1 – cos2θ = sin2θ]
= R.H.S.
