Advertisements
Advertisements
Question
Find the value of sin 30° + cos 60°.
Advertisements
Solution
sin 30° + cos 60° = `1/2 + 1/2`
= `(1 + 1)/2`
= `2/2`
= 1
∴ sin 30° + cos 60° = 1
APPEARS IN
RELATED QUESTIONS
If m=(acosθ + bsinθ) and n=(asinθ – bcosθ) prove that m2+n2=a2+b2
As observed from the top of an 80 m tall lighthouse, the angles of depression of two ships on the same side of the lighthouse of the horizontal line with its base are 30° and 40° respectively. Find the distance between the two ships. Give your answer correct to the nearest meter.
Prove the following trigonometric identities
`(1 + tan^2 theta)/(1 + cot^2 theta) = ((1 - tan theta)/(1 - cot theta))^2 = tan^2 theta`
Prove the following trigonometric identities.
`(cot A - cos A)/(cot A + cos A) = (cosec A - 1)/(cosec A + 1)`
Prove the following trigonometric identities.
(cosec θ − sec θ) (cot θ − tan θ) = (cosec θ + sec θ) ( sec θ cosec θ − 2)
Prove the following trigonometric identities.
`tan A/(1 + tan^2 A)^2 + cot A/((1 + cot^2 A)) = sin A cos A`
Prove the following identities:
sec2 A . cosec2 A = tan2 A + cot2 A + 2
Prove the following identities:
`1/(cosA + sinA) + 1/(cosA - sinA) = (2cosA)/(2cos^2A - 1)`
Prove the following identities:
`(1 + (secA - tanA)^2)/(cosecA(secA - tanA)) = 2tanA`
`(1 + cot^2 theta ) sin^2 theta =1`
`1/((1+tan^2 theta)) + 1/((1+ tan^2 theta))`
`sqrt((1 + sin θ)/(1 - sin θ)) = sec θ + tan θ`
Write the value of sin A cos (90° − A) + cos A sin (90° − A).
Write True' or False' and justify your answer the following :
The value of the expression \[\sin {80}^° - \cos {80}^°\]
Prove the following identity :
( 1 + cotθ - cosecθ) ( 1 + tanθ + secθ)
Prove the following identity :
`sin^4A + cos^4A = 1 - 2sin^2Acos^2A`
Prove that sin4θ - cos4θ = sin2θ - cos2θ
= 2sin2θ - 1
= 1 - 2 cos2θ
Prove that: `sqrt((1 - cos θ)/(1 + cos θ)) = "cosec" θ - cot θ`.
Given that sinθ + 2cosθ = 1, then prove that 2sinθ – cosθ = 2.
Complete the following activity to prove:
cotθ + tanθ = cosecθ × secθ
Activity: L.H.S. = cotθ + tanθ
= `cosθ/sinθ + square/cosθ`
= `(square + sin^2theta)/(sinθ xx cosθ)`
= `1/(sinθ xx cosθ)` ....... ∵ `square`
= `1/sinθ xx 1/cosθ`
= `square xx secθ`
∴ L.H.S. = R.H.S.
