Advertisements
Advertisements
प्रश्न
tan (90 – θ) = ?
पर्याय
sin θ
cos θ
cot θ
tan θ
Advertisements
उत्तर
cot θ
Explanation:
tan(90° – θ)
= `sin(90^circ - θ)/cos(90^circ - θ)`
= `(cos θ)/(sin θ)`
= cot θ
APPEARS IN
संबंधित प्रश्न
If m=(acosθ + bsinθ) and n=(asinθ – bcosθ) prove that m2+n2=a2+b2
Prove the following identities, where the angles involved are acute angles for which the expressions are defined:
`(cosec θ – cot θ)^2 = (1-cos theta)/(1 + cos theta)`
Prove the following identities, where the angles involved are acute angles for which the expressions are defined:
`(cos A-sinA+1)/(cosA+sinA-1)=cosecA+cotA ` using the identity cosec2 A = 1 cot2 A.
Prove that `(tan^2 theta)/(sec theta - 1)^2 = (1 + cos theta)/(1 - cos theta)`
Prove the following trigonometric identities
sec4 A(1 − sin4 A) − 2 tan2 A = 1
if `a cos^3 theta + 3a cos theta sin^2 theta = m, a sin^3 theta + 3 a cos^2 theta sin theta = n`Prove that `(m + n)^(2/3) + (m - n)^(2/3)`
If a cos θ + b sin θ = m and a sin θ – b cos θ = n, prove that a2 + b2 = m2 + n2
Prove the following identities:
(cos A + sin A)2 + (cos A – sin A)2 = 2
Prove the following identities:
`(1 - cosA)/sinA + sinA/(1 - cosA)= 2cosecA`
` (sin theta - cos theta) / ( sin theta + cos theta ) + ( sin theta + cos theta ) / ( sin theta - cos theta ) = 2/ ((2 sin^2 theta -1))`
`(sec theta + tan theta )/( sec theta - tan theta ) = ( sec theta + tan theta )^2 = 1+2 tan^2 theta + 25 sec theta tan theta `
`(1+ tan theta + cot theta )(sintheta - cos theta) = ((sec theta)/ (cosec^2 theta)-( cosec theta)/(sec^2 theta))`
Write the value of ` sec^2 theta ( 1+ sintheta )(1- sintheta).`
What is the value of \[\sin^2 \theta + \frac{1}{1 + \tan^2 \theta}\]
\[\frac{x^2 - 1}{2x}\] is equal to
Without using trigonometric table , evaluate :
`sin72^circ/cos18^circ - sec32^circ/(cosec58^circ)`
Find the value of `θ(0^circ < θ < 90^circ)` if :
`cos 63^circ sec(90^circ - θ) = 1`
For ΔABC , prove that :
`tan ((B + C)/2) = cot "A/2`
Without using trigonometric identity , show that :
`sec70^circ sin20^circ - cos20^circ cosec70^circ = 0`
`1/sin^2θ - 1/cos^2θ - 1/tan^2θ - 1/cot^2θ - 1/sec^2θ - 1/("cosec"^2θ) = -3`, then find the value of θ.
