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प्रश्न
Prove that `((tan 20°)/(cosec 70°))^2 + ((cot 20°)/(sec 70°))^2 = 1`
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उत्तर
LHS = `((tan 20°)/(cosec 70°))^2 + ((cot 20°)/(sec 70°))^2`
= `((sin 20°. sin 70°)/(cos 20°))^2 + ((cos 20°. cos 20°)/(sin 20°))^2`
= `[(sin 20°.sin (90° - 20°))/(cos 20°)]^2 + [(cos 20°. cos(90° - 20°))/(sin 20°)]^2`
= `[ (sin 20°.cos 20°)/(cos 20°)]^2 + [(cos 20°. sin 20°)/(sin 20°)]^2`
= sin2 20° + cos2 20°
= 1
= RHS
Hence proved.
संबंधित प्रश्न
(1 + tan θ + sec θ) (1 + cot θ − cosec θ) = ______.
`(1+tan^2A)/(1+cot^2A)` = ______.
Prove that `(sin theta)/(1-cottheta) + (cos theta)/(1 - tan theta) = cos theta + sin theta`
Prove the following trigonometric identities
`(1 + tan^2 theta)/(1 + cot^2 theta) = ((1 - tan theta)/(1 - cot theta))^2 = tan^2 theta`
`(1 + cot^2 theta ) sin^2 theta =1`
`1/((1+ sin θ)) + 1/((1 - sin θ)) = 2 sec^2 θ`
`(1+tan^2theta)(1+cot^2 theta)=1/((sin^2 theta- sin^4theta))`
`(sin theta)/((sec theta + tan theta -1)) + cos theta/((cosec theta + cot theta -1))=1`
sin4A – cos4A = 1 – 2cos2A. For proof of this complete the activity given below.
Activity:
L.H.S. = `square`
= (sin2A + cos2A) `(square)`
= `1 (square)` ...`[sin^2"A" + square = 1]`
= `square` – cos2A ...[sin2A = 1 – cos2A]
= `square`
= R.H.S.
Prove the following trigonometry identity:
(sin θ + cos θ)(cosec θ – sec θ) = cosec θ ⋅ sec θ – 2 tan θ
