Advertisements
Advertisements
प्रश्न
Prove the following identity :
`(cosecA - sinA)(secA - cosA) = 1/(tanA + cotA)`
उत्तर
LHS = `(cosecA - sinA)(secA - cosA)`
= `(1/sinA - sinA)(1/cosA - cosA)`
= `((1-sin^2A)/(sinA))((1 - cos^2A)/cosA)`
= `(cos^2A/sinA)(sin^2A/cosA)` = cosA.sinA
RHS = `1/(tanA + cotA)`
= `1/(sinA/cosA + cosA/sinA) = 1/((sin^2A + cos^2A)/(sinA.cosA))` = cosA.sinA
Hence , LHS = RHS
APPEARS IN
संबंधित प्रश्न
Prove that `(sin theta)/(1-cottheta) + (cos theta)/(1 - tan theta) = cos theta + sin theta`
Prove the following trigonometric identities.
`cot theta - tan theta = (2 cos^2 theta - 1)/(sin theta cos theta)`
Prove that:
(sec A − tan A)2 (1 + sin A) = (1 − sin A)
`(1 + cot^2 theta ) sin^2 theta =1`
`(sec^2 theta-1) cot ^2 theta=1`
Find the value of `(cos 38° cosec 52°)/(tan 18° tan 35° tan 60° tan 72° tan 55°)`
What is the value of \[6 \tan^2 \theta - \frac{6}{\cos^2 \theta}\]
Prove that `sqrt((1 - sin θ)/(1 + sin θ)) = sec θ - tan θ`.
If 4 tanβ = 3, then `(4sinbeta-3cosbeta)/(4sinbeta+3cosbeta)=` ______.
Statement 1: sin2θ + cos2θ = 1
Statement 2: cosec2θ + cot2θ = 1
Which of the following is valid?
