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प्रश्न
Prove that cot θ. tan (90° - θ) - sec (90° - θ). cosec θ + 1 = 0.
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उत्तर
LHS = cot θ. tan (90° - θ) - sec (90° - θ). cosec θ + 1
= cot θ. cot θ - cosec θ. cosec θ + 1
= (cot2θ - cosec2θ) + 1
= - 1 + 1 = 0
= RHS
Hence proved.
संबंधित प्रश्न
Prove the following trigonometric identities.
`(cos^2 theta)/sin theta - cosec theta + sin theta = 0`
Prove the following trigonometric identities.
`(1 + sin θ)/cos θ+ cos θ/(1 + sin θ) = 2 sec θ`
Prove the following trigonometric identities.
`(tan A + tan B)/(cot A + cot B) = tan A tan B`
`sin^2 theta + 1/((1+tan^2 theta))=1`
`(tan theta)/((sec theta -1))+(tan theta)/((sec theta +1)) = 2 sec theta`
If `( cosec theta + cot theta ) =m and ( cosec theta - cot theta ) = n, ` show that mn = 1.
Simplify
sin A `[[sinA -cosA],["cos A" " sinA"]] + cos A[[ cos A" sin A " ],[-sin A" cos A"]]`
Prove that: 2(sin6 θ + cos6 θ) – 3 (sin4 θ + cos4 θ) + 1 = 0.
If `tan θ = 9/40`, complete the activity to find the value of sec θ.
Activity:
sec2θ = 1 + `square` ...[Fundamental trigonometric identity]
sec2θ = 1 + `square^2`
sec2θ = 1 + `square`
sec θ = `square`
To prove cot θ + tan θ = cosec θ × sec θ, complete the activity given below.
Activity:
L.H.S. = `square`
= `square/(sinθ) + (sinθ)/(cosθ)`
= `(cos^2θ + sin^2θ)/square`
= `1/(sinθ.cosθ)` ...`[cos^2θ + sin^2θ = square]`
= `1/(sinθ) xx 1/square`
= `square`
= R.H.S.
