Advertisements
Advertisements
प्रश्न
Prove that:
`(cos^3 θ + sin^3 θ)/(cos θ + sin θ) + (cos^3 θ - sin^3 θ)/(cos θ - sin θ) = 2`
Advertisements
उत्तर
LHS = `(cos^3 θ + sin^3 θ)/(cos θ + sin θ) + (cos^3 θ - sin^3 θ)/(cos θ - sin θ)`
= `((cos θ + sin θ)(cos^2 θ + sin^2 θ - cos θ sin θ))/(cos θ + sin θ) + ((cos θ - sin θ)(cos^2 θ + sin^2 θ - cos θ sin θ))/(cos θ - sin θ)`
= 1 - sin θ cos θ + 1 + sin θ cos θ
= 2
= RHS
Hence proved.
संबंधित प्रश्न
Write the value of ` sec^2 theta ( 1+ sintheta )(1- sintheta).`
If \[\sin \theta = \frac{1}{3}\] then find the value of 2cot2 θ + 2.
Prove the following identity :
`(1 - sin^2θ)sec^2θ = 1`
If tan θ = 2, where θ is an acute angle, find the value of cos θ.
Prove that (cosec A - sin A)( sec A - cos A) sec2 A = tan A.
Without using trigonometric table, prove that
`cos^2 26° + cos 64° sin 26° + (tan 36°)/(cot 54°) = 2`
Prove the following identities.
`sqrt((1 + sin theta)/(1 - sin theta)` = sec θ + tan θ
a cot θ + b cosec θ = p and b cot θ + a cosec θ = q then p2 – q2 is equal to
If tan θ × A = sin θ, then A = ?
Show that: `tan "A"/(1 + tan^2 "A")^2 + cot "A"/(1 + cot^2 "A")^2 = sin"A" xx cos"A"`
