Advertisements
Advertisements
Question
If secθ + tanθ = m , secθ - tanθ = n , prove that mn = 1
Advertisements
Solution
LHS = mn = (secθ + tanθ) (secθ - tanθ)
⇒ LHS = `sec^2θ - tan^2θ` [Because (a-b)(a+b) = a2 - b2]
⇒ LHS = 1 [Since `1 + tan^2θ = sec^2θ`]
Hence , mn = 1
APPEARS IN
RELATED QUESTIONS
Prove the following identities:
`1/(1 - sinA) + 1/(1 + sinA) = 2sec^2A`
`sec theta (1- sin theta )( sec theta + tan theta )=1`
What is the value of (1 − cos2 θ) cosec2 θ?
Prove the following identity :
`(sec^2θ - sin^2θ)/tan^2θ = cosec^2θ - cos^2θ`
Without using trigonometric identity , show that :
`cos^2 25^circ + cos^2 65^circ = 1`
Prove that sin4θ - cos4θ = sin2θ - cos2θ
= 2sin2θ - 1
= 1 - 2 cos2θ
Prove the following identities:
`1/(sin θ + cos θ) + 1/(sin θ - cos θ) = (2sin θ)/(1 - 2 cos^2 θ)`.
Prove that `(1 + tan^2 A)/(1 + cot^2 A)` = sec2 A – 1
Find the value of sin2θ + cos2θ

Solution:
In Δ ABC, ∠ABC = 90°, ∠C = θ°
AB2 + BC2 = `square` .....(Pythagoras theorem)
Divide both sides by AC2
`"AB"^2/"AC"^2 + "BC"^2/"AC"^2 = "AC"^2/"AC"^2`
∴ `("AB"^2/"AC"^2) + ("BC"^2/"AC"^2) = 1`
But `"AB"/"AC" = square and "BC"/"AC" = square`
∴ `sin^2 theta + cos^2 theta = square`
Prove the following trigonometry identity:
(sin θ + cos θ)(cosec θ – sec θ) = cosec θ ⋅ sec θ – 2 tan θ
