Question

sec70 sin20 – cos20 cosec70,Evaluate it.

Answer 1

sec70°. sin20° - cos20° .cosec70°

we know, secx = 1/cosx and cosecx = 1/sinx

so, 1/cos70° × sin20° - cos20° × 1/sin70°

= sin20°/cos70° - cos20°/sin70°

we know, sin(90 - x) = cosx
so, sin20° = sin(90° - 70°) = cos70° ---(1)

again, cos(90 - x) = sinx
so, cos20° = cos(90° - 20°) = sin70° ---(2)

now, from equations (1) and (2),

= cos70°/cos70° - sin70°/sin70°

= 1 - 1 = 0

hence, answer is 0

Answer 2

HELLO DEAR,

GIVEN:-
sec70°. sin20° - cos20° .cosec70°

we know:-
secx = 1/cosx and cosecx = 1/sinx
sin(90 - x) = cosx


so, 1/cos70° * sin20° - cos20° * 1/sin70°

= sin20°/cos70° - cos20°/sin70°

[cos(90 - x) = sinx ]

= cos70°/cos70° - sin70°/sin70°

1 - 1 = 0


I HOPE ITS HELP YOU DEAR,
THANKS