Advertisements
Advertisements
Question
If tan θ – sin2θ = cos2θ, then show that sin2 θ = `1/2`.
Advertisements
Solution
tan θ – sin2θ = cos2θ ......[Given]
∴ tan θ = sin2θ + cos2θ
∴ tan θ = 1 ....[∵ sin2θ + cos2θ = 1]
But, tan 45° = 1
∴ tan θ = tan 45°
∴ θ = 45°
sin2θ = sin245°
= `(1/sqrt(2))^2`
= `1/2`
APPEARS IN
RELATED QUESTIONS
Prove the following trigonometric identities.
`cos theta/(1 + sin theta) = (1 - sin theta)/cos theta`
Prove the following trigonometric identities.
`tan^2 theta - sin^2 theta tan^2 theta sin^2 theta`
Prove the following trigonometric identities.
`(1 + cos A)/sin A = sin A/(1 - cos A)`
Prove the following trigonometric identities.
(cosec θ − sec θ) (cot θ − tan θ) = (cosec θ + sec θ) ( sec θ cosec θ − 2)
Write the value of `(1 + tan^2 theta ) cos^2 theta`.
Write the value of cos1° cos 2°........cos180° .
If 5x = sec ` theta and 5/x = tan theta , " find the value of 5 "( x^2 - 1/( x^2))`
Four alternative answers for the following question are given. Choose the correct alternative and write its alphabet:
sin θ × cosec θ = ______
What is the value of 9cot2 θ − 9cosec2 θ?
Prove the following identity :
( 1 + cotθ - cosecθ) ( 1 + tanθ + secθ)
Prove the following identity :
`(cos^3θ + sin^3θ)/(cosθ + sinθ) + (cos^3θ - sin^3θ)/(cosθ - sinθ) = 2`
If `x/(a cosθ) = y/(b sinθ) "and" (ax)/cosθ - (by)/sinθ = a^2 - b^2 , "prove that" x^2/a^2 + y^2/b^2 = 1`
Evaluate:
sin2 34° + sin2 56° + 2 tan 18° tan 72° – cot2 30°
Find the value of ( sin2 33° + sin2 57°).
Prove that: `cos^2 A + 1/(1 + cot^2 A) = 1`.
If tan θ = `9/40`, complete the activity to find the value of sec θ.
Activity:
sec2θ = 1 + `square` ......[Fundamental trigonometric identity]
sec2θ = 1 + `square^2`
sec2θ = 1 + `square`
sec θ = `square`
Prove that cosec θ – cot θ = `sin theta/(1 + cos theta)`
If sin θ + cos θ = `sqrt(3)`, then show that tan θ + cot θ = 1
Show that, cotθ + tanθ = cosecθ × secθ
Solution :
L.H.S. = cotθ + tanθ
= `cosθ/sinθ + sinθ/cosθ`
= `(square + square)/(sinθ xx cosθ)`
= `1/(sinθ xx cosθ)` ............... `square`
= `1/sinθ xx 1/square`
= cosecθ × secθ
L.H.S. = R.H.S
∴ cotθ + tanθ = cosecθ × secθ
Prove the following identity:
(sin2θ – 1)(tan2θ + 1) + 1 = 0
