Question

Cosec^2θ-1/cosec^2θ=cos^2θ​

Answer 1

see explanation

Explanation:

Using the identities

cosec x =
`(1)/(sinx)` , 1 - sin²θ = 1

Consider the left side

`(cosec^20-1)/(cosec^20)`

=
`((1)/(sin^20)-(sin^20)/(sin^20) )/((1)/(sin^20) )`

= (
`(1)/(sin^20)-(sin^20)/(sin^20)` ) ×
`(sin^20)/(1)` ← distribute parenthesis by sin²θ

= 1 - sin²θ

= cos²θ

= right side, thus proven

Answer 2

cosec(2t)=1/sin(2t) ==> 1/(2×cos(t)×sin(t))

cos(2t)= 1-2sin²t

or 2cos²t-1

or cos²t - sin²t

after we knew this informations we have to write it in the question

1/2×sint×cost-1/1/2×sint×cost=cos2t

-2×sint×cost+1/2sint×cost/1/2sint×cost=cos2t

the 2sint×cost will gone

-2×sint×cost+1=cos2t

lets take cos2t =1-2sin²t

-2×sint×cost+1=1-2sin²t

-2sint×cost=-2sin²t

cost=sint

that means the t is equal to 45 because sin45 and cos45 equal to √2/2

hopefully i didn't waste your time by reading and not understanding my English is not that good sorry