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प्रश्न
Without using trigonometric table , evaluate :
`sin72^circ/cos18^circ - sec32^circ/(cosec58^circ)`
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उत्तर
`sin72^circ/cos18^circ - sec32^circ/(cosec58^circ)`
⇒ `sin(90^circ - 18^circ)/cos18^circ - sec(90^circ - 58^circ)/(cosec58^circ)`
⇒ `cos18^circ/cos18^circ - (cosec 58^circ)/(cosec58^circ) = 1 - 1 = 0`
APPEARS IN
संबंधित प्रश्न
`"If "\frac{\cos \alpha }{\cos \beta }=m\text{ and }\frac{\cos \alpha }{\sin \beta }=n " show that " (m^2 + n^2 ) cos^2 β = n^2`
Prove the following trigonometric identities.
`(1 - cos theta)/sin theta = sin theta/(1 + cos theta)`
Prove the following identities:
`(1 + sin A)/(1 - sin A) = (cosec A + 1)/(cosec A - 1)`
Prove the following identities:
(1 + tan A + sec A) (1 + cot A – cosec A) = 2
Prove the following identity :
`(1 + cosA)/(1 - cosA) = (cosecA + cotA)^2`
Prove that sin (90° - θ) cos (90° - θ) = tan θ. cos2θ.
Prove the following identities.
`(1 - tan^2theta)/(cot^2 theta - 1)` = tan2 θ
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.
If sinA + sin2A = 1, then the value of the expression (cos2A + cos4A) is ______.
Prove the following:
`tanA/(1 + sec A) - tanA/(1 - sec A)` = 2cosec A
