Question

help me in this question

Answer 1

LHS = [sin70° / cos 20°] + [cosec20° / sec70°] -
2 cos70°cosec 20°

= [sin(90-20) / cos20°] +[cosec(90-70) /
sec70°] - 2cos70° (1/ sin(90-70) )

= [cos20° / cos20°] + [sec70° / sec70°] -
2cos70°[1/ cos 70°]

= 1 + 1 - 2

= 2 - 2

= 0

= RHS

So, LHS = RHS
Hence proved.

Answer 2

AS WE KNOW

SIN(THETA)=COS(90-THETA)

AND

COSEC(THETA)

=SEC(90-THETA)

AND

SIN(THETA)

=1/ COSEC(THETA)

SO IN THIS QUESTION

SIN70°=COS(90-70)°

=COS20°

AND

COSEC20°=SEC(90-20°)

=SEC(70°)

AND

2COS70°=2SIN20°

AND COSEC20°=1/SIN20°

SO

IT IS

(COS20°/COS20°)+

(SEC70°/SEC70°)+

2SIN20°×(1/SIN20°)

=1+1-2=0

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