Advertisements
Advertisements
Question
`1+ (cot^2 theta)/((1+ cosec theta))= cosec theta`
Advertisements
Solution
LHS =` 1+(cot^2 theta)/((1+ cosectheta))`
=`1+((cosec^2 theta-1))/((cosectheta++1)) (∵ cosec^2 theta - cot^2 theta =1)`
=`1+((cosectheta+1)(cosec theta-1))/((cosec theta +1))`
=`1+ (cosec theta -1)`
=` cosec theta`
=RHS
RELATED QUESTIONS
Evaluate
`(sin ^2 63^@ + sin^2 27^@)/(cos^2 17^@+cos^2 73^@)`
Prove the following identities:
`sqrt((1 + sinA)/(1 - sinA)) = sec A + tan A`
Prove the following identities:
`(costhetacottheta)/(1 + sintheta) = cosectheta - 1`
Prove the following identities:
`1 - sin^2A/(1 + cosA) = cosA`
`1 + (tan^2 θ)/((1 + sec θ)) = sec θ`
If `tan theta = 1/sqrt(5), "write the value of" (( cosec^2 theta - sec^2 theta))/(( cosec^2 theta - sec^2 theta))`.
Write True' or False' and justify your answer the following :
The value of \[\sin \theta\] is \[x + \frac{1}{x}\] where 'x' is a positive real number .
Prove the following identity :
`sin^2Acos^2B - cos^2Asin^2B = sin^2A - sin^2B`
Prove the following identity :
`sec^2A.cosec^2A = tan^2A + cot^2A + 2`
Prove the following identity :
`(cos^3θ + sin^3θ)/(cosθ + sinθ) + (cos^3θ - sin^3θ)/(cosθ - sinθ) = 2`
Without using trigonometric table , evaluate :
`sin72^circ/cos18^circ - sec32^circ/(cosec58^circ)`
Without using trigonometric identity , show that :
`sec70^circ sin20^circ - cos20^circ cosec70^circ = 0`
If (sin α + cosec α)2 + (cos α + sec α)2 = k + tan2α + cot2α, then the value of k is equal to
Choose the correct alternative:
sec 60° = ?
If tan θ = `13/12`, then cot θ = ?
`5/(sin^2theta) - 5cot^2theta`, complete the activity given below.
Activity:
`5/(sin^2theta) - 5cot^2theta`
= `square (1/(sin^2theta) - cot^2theta)`
= `5(square - cot^2theta) ......[1/(sin^2theta) = square]`
= 5(1)
= `square`
tan2θ – sin2θ = tan2θ × sin2θ. For proof of this complete the activity given below.
Activity:
L.H.S = `square`
= `square (1 - (sin^2theta)/(tan^2theta))`
= `tan^2theta (1 - square/((sin^2theta)/(cos^2theta)))`
= `tan^2theta (1 - (sin^2theta)/1 xx (cos^2theta)/square)`
= `tan^2theta (1 - square)`
= `tan^2theta xx square` .....[1 – cos2θ = sin2θ]
= R.H.S
Prove that `"cot A"/(1 - tan "A") + "tan A"/(1 - cot"A")` = 1 + tan A + cot A = sec A . cosec A + 1
Prove the following that:
`tan^3θ/(1 + tan^2θ) + cot^3θ/(1 + cot^2θ)` = secθ cosecθ – 2 sinθ cosθ
