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प्रश्न
Prove that : `(sin(90° - θ) tan(90° - θ) sec (90° - θ))/(cosec θ. cos θ. cot θ) = 1`
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उत्तर
LHS = `(sin(90° - θ) tan(90° - θ) sec (90° - θ))/(cosec θ. cos θ. cot θ) = 1`
= `(cosec θ. cos θ. cot θ)/(cosec θ. cos θ. cot θ)`
= 1
= RHS
Hence proved.
संबंधित प्रश्न
Prove the following trigonometric identities:
(1 – cos2 A) cosec2 A = 1
Prove the following trigonometric identities.
`[tan θ + 1/cos θ]^2 + [tan θ - 1/cos θ]^2 = 2((1 + sin^2 θ)/(1 - sin^2 θ))`
Prove the following identities:
`(1 - cosA)/sinA + sinA/(1 - cosA)= 2cosecA`
Prove the following identities:
sec4 A (1 – sin4 A) – 2 tan2 A = 1
`1+ (cot^2 theta)/((1+ cosec theta))= cosec theta`
Prove the following identity :
`(tanθ + sinθ)/(tanθ - sinθ) = (secθ + 1)/(secθ - 1)`
Prove that `(cot "A" + "cosec A" - 1)/(cot "A" - "cosec A" + 1) = (1 + cos "A")/sin "A"`
Prove the following identities.
sec4 θ (1 – sin4 θ) – 2 tan2 θ = 1
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.
Which of the following is true for all values of θ (0° ≤ θ ≤ 90°)?
