Question

√sec²Ф+cosec²Ф = tanФ+cotФ
it is fully sq root of LHS.

Answer 1

LHS;
=>    √1/cos²Ф +1/sin²Ф
=>    √(sin²Ф+cos²Ф)/(sin²Фcos²Ф)
=>     √1/(sin²Фcos²Ф)
=>       1/sinФcosФ

RHS;
=>   tanФ+cotФ
=>   sinФ/cosФ+cosФ/sinФ
=>   (sin²Ф+cos²Ф)/sinФcosФ
=>   1/sinФcosФ

Thus LHS=RHS,

Hence Proved...

Answer 2

sec^2Φ=1/cos^2Φ
cosec^2Φ=1/sin^2Φ
adding these two we get sin^2Φ +cos^2Φ/sin^2Φ cos^2Φ
therefore the answer is 1/sinΦcosΦ
on the otherhand RHS is tanΦ+cotΦ=sinΦ/cosΦ+cosΦ/sinΦ=1/sinΦcosΦ