Question

Find the value of 0 (0º < < 90°) if:
(1) cos 63° sec (90° - 0) = 1
(ii) tan 35° cot (90° - 0) = 1.​

Answer 1

i) For the first equation we can take the value of O as 27°

Because,

cos 63° × sec (90°-27°) = 1

sin 27° × cosec 27° = 1

sin 27° × 1/sin 27° = 1

So, 1 = 1

We can write cos 63° as sin 27° as it is complementary.And we can write

cot(90°-27°) as cosec27° which can also be written as 1/sin27°.

ii) For the second equation we can take the value of O as 55°

Because,

tan 35° × cot (90° - 55°) = 1

tan35° × cot 35° = 1

tan35° × 1/tan35° = 1

So, 1 = 1

Here we r not taking anything as complementary

Instead taking 90° - 55° as 35° only so that we can equate both.

*Hope this helps U*