Question

show that [cos^2(45+thita) + cos^2(45-thita)]/[tan(60+thita)tan(30-thita)] =1​

Answer 1

Step-by-step explanation:

when A+B=90°

cos²A+cos²B=1 --> (1)

[proof: cos²A+cos²B

cos²A+cos²(90-A)

cos²A+sin²A

=1]

A+B

(45+thita)+(45-thita)=90°

so,cos^2(45+thita) + cos^2(45-thita)=1

tan(thita)= cot(90-thita)

tan(30-thita)=cot(90-(30-thita))

=cot(60+thita)

=1/tan(60+thita) -->(2)

consider LHS

[cos^2(45+thita) + cos^2(45-thita)]/[tan(60+thita)tan(30-thita)]

from (1) and (2)

1/1

=1(R.H.S)