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प्रश्न
Without using trigonometric identity , show that :
`sec70^circ sin20^circ - cos20^circ cosec70^circ = 0`
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उत्तर
`sec70^circ sin20^circ - cos20^circ cosec70^circ = 0`
Consider `sec70^circ sin20^circ - cos20^circ cosec70^circ`
⇒ `sec(90^circ - 20^circ)sin20^circ - cos20^circ . cosec(90^circ - 20^circ)`
⇒ `cosec20^circ sin20^circ - cos20^circ sec20^circ`
⇒ `1/sin20^circ . sin20^circ - cos20^circ . 1/cos20^circ`
⇒ 1 - 1 = 0
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संबंधित प्रश्न
Prove the following trigonometric identities.
`((1 + tan^2 theta)cot theta)/(cosec^2 theta) = tan theta`
`(sin theta)/((sec theta + tan theta -1)) + cos theta/((cosec theta + cot theta -1))=1`
`(tan A + tanB )/(cot A + cot B) = tan A tan B`
If (cot θ + tan θ) = m and (sec θ – cos θ) = n, prove that `(m^2 n)^(2//3) - (mn^2)^(2//3) = 1`.
Prove the following identity :
`(1 + sinA)/(1 - sinA) = (cosecA + 1)/(cosecA - 1)`
Prove the following identity :
`tanA - cotA = (1 - 2cos^2A)/(sinAcosA)`
Prove that sin (90° - θ) cos (90° - θ) = tan θ. cos2θ.
Prove that: `cos^2 A + 1/(1 + cot^2 A) = 1`.
Prove that sec2θ + cosec2θ = sec2θ × cosec2θ.
Prove that `sqrt(sec^2 theta + "cosec"^2 theta) = tan theta + cot theta`
