Advertisements
Advertisements
Question
If x = a sec θ + b tan θ and y = a tan θ + b sec θ prove that x2 - y2 = a2 - b2.
Advertisements
Solution
Here,
x2 = a2 sec2θ + 2ab sec θ.tan θ + b2tan2θ
y2 = a2 tan2θ + 2ab sec θ.tan θ + b2sec2θ
⇒ x2 - y2 = a2 ( sec2θ - tan2θ ) - b2 ( sec2θ - tan2θ )
⇒ x2 - y2 = a2 - b2. ....( ∵ sec2θ - tan2θ = 1)
Hence proved.
RELATED QUESTIONS
Prove that:
`(cosecA - sinA)(secA - cosA) = 1/(tanA + cotA)`
If (cosec θ – sin θ) = a3 and (sec θ – cos θ) = b3, prove that a2b2(a2 + b2) = 1.
If `cos theta = 7/25 , "write the value of" ( tan theta + cot theta).`
If `tan theta = 1/sqrt(5), "write the value of" (( cosec^2 theta - sec^2 theta))/(( cosec^2 theta - sec^2 theta))`.
If `sin theta = x , " write the value of cot "theta .`
Prove that:
`"tanθ"/("secθ" – 1) = (tanθ + secθ + 1)/(tanθ + secθ - 1)`
Prove the following identity :
secA(1 - sinA)(secA + tanA) = 1
Prove the following identity :
`(sinA - sinB)/(cosA + cosB) + (cosA - cosB)/(sinA + sinB) = 0`
tan2θ – sin2θ = tan2θ × sin2θ. For proof of this complete the activity given below.
Activity:
L.H.S. = `square`
= `square (1 - (sin^2θ)/(tan^2θ))`
= `tan^2θ (1 - square/((sin^2θ)/(cos^2θ)))`
= `tan^2θ (1 - (sin^2θ)/1 xx (cos^2θ)/square)`
= `tan^2θ (1 - square)`
= `tan^2θ xx square` ...[1 – cos2θ = sin2θ]
= R.H.S.
The value of tan A + sin A = M and tan A - sin A = N.
The value of `("M"^2 - "N"^2) /("MN")^0.5`
