Advertisements
Advertisements
Question
Show that `(cos^2(45^circ + θ) + cos^2(45^circ - θ))/(tan(60^circ + θ) tan(30^circ - θ)) = 1`
Advertisements
Solution
L.H.S = `(cos^2(45^circ + θ) + cos^2(45^circ - θ))/(tan(60^circ + θ) xx tan(30^circ - θ))`
= `(cos^2(45^circ + θ) + [sin{90^circ - (45^circ - θ)}]^2)/(tan(60^circ + θ) xx cot{90^circ - (30^circ - θ)})` ...[∵ sin (90° – θ) = cos θ and cot (90° – θ) = tan θ]
= `(cos^2(45^circ + θ) + sin^2(45^circ + θ))/(tan(60^circ + θ) xx cot(60^circ + θ))` ...[∵ sin2θ + cos2θ = 1]
= `1/(tan(60^circ + θ)) xx 1/(tan(60^circ + θ))` ...`[∵ cot θ = 1/tan θ]`
= 1
= R.H.S
RELATED QUESTIONS
Prove that `(sin theta)/(1-cottheta) + (cos theta)/(1 - tan theta) = cos theta + sin theta`
Prove the following trigonometric identities.
`(1 - cos theta)/sin theta = sin theta/(1 + cos theta)`
Prove the following identities:
`cot^2A/(cosecA + 1)^2 = (1 - sinA)/(1 + sinA)`
Prove the following identities:
`(cotA - cosecA)^2 = (1 - cosA)/(1 + cosA)`
Prove that:
`cot^2A/(cosecA - 1) - 1 = cosecA`
Write the value of `(1+ tan^2 theta ) ( 1+ sin theta ) ( 1- sin theta)`
Prove that:
`(sin^2θ)/(cosθ) + cosθ = secθ`
Define an identity.
If sin2 θ cos2 θ (1 + tan2 θ) (1 + cot2 θ) = λ, then find the value of λ.
Write True' or False' and justify your answer the following :
The value of sin θ+cos θ is always greater than 1 .
The value of sin2 29° + sin2 61° is
Prove the following identity :
`(1 + cotA)^2 + (1 - cotA)^2 = 2cosec^2A`
Prove the following identity :
`cosA/(1 - tanA) + sin^2A/(sinA - cosA) = cosA + sinA`
Verify that the points A(–2, 2), B(2, 2) and C(2, 7) are the vertices of a right-angled triangle.
Prove that: `cos^2 A + 1/(1 + cot^2 A) = 1`.
Prove that `(tan θ + sin θ)/(tan θ - sin θ) = (sec θ + 1)/(sec θ - 1)`
Prove the following identities.
`(sin "A" - sin "B")/(cos "A" + cos "B") + (cos "A" - cos "B")/(sin "A" + sin "B")`
If cos A = `(2sqrt(m))/(m + 1)`, then prove that cosec A = `(m + 1)/(m - 1)`.
Prove that 2(sin6A + cos6A) – 3(sin4A + cos4A) + 1 = 0.
tan θ × `sqrt(1 - sin^2 θ)` is equal to:
