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प्रश्न
If tan θ = `7/24`, then to find value of cos θ complete the activity given below.
Activity:
sec2θ = 1 + `square` ......[Fundamental tri. identity]
sec2θ = 1 + `square^2`
sec2θ = 1 + `square/576`
sec2θ = `square/576`
sec θ = `square`
cos θ = `square` .......`[cos theta = 1/sectheta]`
उत्तर
sec2θ = 1 + tan2θ ......[Fundamental tri. identity]
∴ sec2θ = 1 + `(7/24)^2`
∴ sec2θ = 1 + `49/576`
∴ sec2θ =`(576 + 49)/576`
∴ sec2θ = `625/576`
∴ sec θ = `25/24`
∴ cos θ = `24/25` .......`[cos theta = 1/sectheta]`
संबंधित प्रश्न
Prove the following identities, where the angles involved are acute angles for which the expressions are defined:
`cos A/(1 + sin A) + (1 + sin A)/cos A = 2 sec A`
The angles of depression of two ships A and B as observed from the top of a light house 60 m high are 60° and 45° respectively. If the two ships are on the opposite sides of the light house, find the distance between the two ships. Give your answer correct to the nearest whole number.
Prove that (cosec A – sin A)(sec A – cos A) sec2 A = tan A.
Prove the following trigonometric identities.
`cot theta - tan theta = (2 cos^2 theta - 1)/(sin theta cos theta)`
Prove the following trigonometric identities.
`(cos theta - sin theta + 1)/(cos theta + sin theta - 1) = cosec theta + cot theta`
Prove the following trigonometric identities.
(1 + cot A − cosec A) (1 + tan A + sec A) = 2
Prove the following identities:
`(sinA + cosA)/(sinA - cosA) + (sinA - cosA)/(sinA + cosA) = 2/(2sin^2A - 1)`
`(sec^2 theta-1) cot ^2 theta=1`
`cos^2 theta + 1/((1+ cot^2 theta )) =1`
`(sin theta +cos theta )/(sin theta - cos theta)+(sin theta- cos theta)/(sin theta + cos theta) = 2/((sin^2 theta - cos ^2 theta)) = 2/((2 sin^2 theta -1))`
Write the value of `(1 + cot^2 theta ) sin^2 theta`.
Prove the following identity :
secA(1 - sinA)(secA + tanA) = 1
Prove the following identity :
`(cotA + cosecA - 1)/(cotA - cosecA + 1) = (cosA + 1)/sinA`
Prove the following identity :
`(1 + tan^2A) + (1 + 1/tan^2A) = 1/(sin^2A - sin^4A)`
If cosθ + sinθ = `sqrt2` cosθ, show that cosθ - sinθ = `sqrt2` sinθ.
Prove the following identities:
`1/(sin θ + cos θ) + 1/(sin θ - cos θ) = (2sin θ)/(1 - 2 cos^2 θ)`.
Prove that `[(1 + sin theta - cos theta)/(1 + sin theta + cos theta)]^2 = (1 - cos theta)/(1 + cos theta)`
Prove that `"cosec" θ xx sqrt(1 - cos^2theta)` = 1
tan2θ – sin2θ = tan2θ × sin2θ. For proof of this complete the activity given below.
Activity:
L.H.S = `square`
= `square (1 - (sin^2theta)/(tan^2theta))`
= `tan^2theta (1 - square/((sin^2theta)/(cos^2theta)))`
= `tan^2theta (1 - (sin^2theta)/1 xx (cos^2theta)/square)`
= `tan^2theta (1 - square)`
= `tan^2theta xx square` .....[1 – cos2θ = sin2θ]
= R.H.S
Factorize: sin3θ + cos3θ
Hence, prove the following identity:
`(sin^3θ + cos^3θ)/(sin θ + cos θ) + sin θ cos θ = 1`
