Question

please solve this question
explain it as well

Answer 1

Answer is given in pic below please report if correct

Answer 2

Given,
tanx = 2
=>tanx = 2/1 = opp side/adj side (in a right angled triangle ABC, where angle A=x, and right angled at B)
AC = hypotenuse, BC = opp side=2, AB = adjacent side=1
From Pythagoras theorem,
(AC)^2 = (AB)^2 +(BC) ^2
(AC)^2 = (1)^2 + (2)^2
(AC)^2 = 1 + 4
(AC)^2 = 5
(AC) = (sq root of 5)

Therefore,
sinx = opp side/hypotenuse = BC/AC = 2/sq root of 5
cosecx = hypotenuse / opp side = AC/BC = sq root of 5/2

We know that,
sinx × sec + tan^2x - cosec x
We also know that
tanx = sinx/ cosx
sinx = tanx(cosx)
So,
tanx(cosx) × secx + tan^2x - cosecx
tanx(1/secx) ×secx + tan^2 - cosecx
"secx gets cancelled"
=> tanx + tan^2 - cosecx
2 + (2)^2 - (sq root of 5)/2
6 - (sq root of 5)/2
(LCM)
Answer : - - - - - [12 - (sq root of 5)] ÷ 2