Advertisements
Advertisements
प्रश्न
Prove that (sin θ + cosec θ)2 + (cos θ + sec θ)2 = 7 + tan2 θ + cot2 θ.
Advertisements
उत्तर १
L.H.S = (sin θ + cosec θ)2 + (cos θ + sec θ)2
= (sin2θ + cosec2θ + 2 sin θ cosec θ + cos2θ + sec2θ + 2cos θ sec θ)
= (sin2θ + cos2θ) + (cosec2θ + sec2θ) + 2 sin θ `(1/("sin"theta)) + 2 cos theta (1/("cos" theta))`
= (1) + (1 + cot2θ + 1 + tan2θ) + (2) + (2)
= 7 + tan2θ + cot2θ
= R.H.S
उत्तर २
L.H.S = (sin θ + cosec θ)2 + (cos θ + sec θ)2
= (sin2θ + cosec2θ + 2 sin θ cosec θ + cos2θ + sec2θ + 2cos θ sec θ)
= (sin2θ + cos2θ ) + 1 + cot2θ + 2 sin θ x `1/sin θ` + 1 + tan2 θ + 2cos θ. `1/cos θ`
= 1 + 1 + 1 + 2 + 2 + tan2 θ + cot2θ
= 7 + tan2 θ + cot2θ
= RHS
Hence proved.
APPEARS IN
संबंधित प्रश्न
Prove the following identities:
`(i) (sinθ + cosecθ)^2 + (cosθ + secθ)^2 = 7 + tan^2 θ + cot^2 θ`
`(ii) (sinθ + secθ)^2 + (cosθ + cosecθ)^2 = (1 + secθ cosecθ)^2`
`(iii) sec^4 θ– sec^2 θ = tan^4 θ + tan^2 θ`
`(1+tan^2A)/(1+cot^2A)` = ______.
Prove the following trigonometric identities.
tan2 A sec2 B − sec2 A tan2 B = tan2 A − tan2 B
Prove the following identities:
`(1 - cosA)/sinA + sinA/(1 - cosA)= 2cosecA`
`cot^2 theta - 1/(sin^2 theta ) = -1`a
If `(cosec theta - sin theta )= a^3 and (sec theta - cos theta ) = b^3 , " prove that " a^2 b^2 ( a^2+ b^2 ) =1`
If `cos theta = 2/3 , " write the value of" (4+4 tan^2 theta).`
If ` cot A= 4/3 and (A+ B) = 90° ` ,what is the value of tan B?
Prove that `(sinθ - cosθ + 1)/(sinθ + cosθ - 1) = 1/(secθ - tanθ)`
If \[\sin \theta = \frac{1}{3}\] then find the value of 2cot2 θ + 2.
\[\frac{\tan \theta}{\sec \theta - 1} + \frac{\tan \theta}{\sec \theta + 1}\] is equal to
If a cos θ + b sin θ = m and a sin θ − b cos θ = n, then a2 + b2 =
Prove the following identity :
`(1 + sinθ)/(cosecθ - cotθ) - (1 - sinθ)/(cosecθ + cotθ) = 2(1 + cotθ)`
If sinA + cosA = m and secA + cosecA = n , prove that n(m2 - 1) = 2m
Prove that:
`(cot A - 1)/(2 - sec^2 A) = cot A/(1 + tan A)`
Evaluate:
`(tan 65^circ)/(cot 25^circ)`
Prove that: sin6θ + cos6θ = 1 - 3sin2θ cos2θ.
`(1 - tan^2 45^circ)/(1 + tan^2 45^circ)` = ?
Prove that (1 – cos2A) . sec2B + tan2B(1 – sin2A) = sin2A + tan2B
Complete the following activity to prove:
cotθ + tanθ = cosecθ × secθ
Activity: L.H.S. = cotθ + tanθ
= `cosθ/sinθ + square/cosθ`
= `(square + sin^2theta)/(sinθ xx cosθ)`
= `1/(sinθ xx cosθ)` ....... ∵ `square`
= `1/sinθ xx 1/cosθ`
= `square xx secθ`
∴ L.H.S. = R.H.S.
