Question

Please ans this fast

Answer 1

The answer is -1
Please refer RD Sharma to get complete solution

Answer 2

lets solve it bit by bit as i cant type whole problem in one go.

lets calculate first part

cos 58 / sin 32

we know for acute angle a

cos a = sin (90 - a)

sin a = cos (90 - a)

so using above formula

cos 58/ sin 32 = sin (90- 58) / sin 32 = sin 32/ sin 32 = 1

similarly 2nd part

sin 22 / cos 68 = cos ( 90- 22)/cos 68 = cos 68/cos 68 = 1

lest solve 3rd part nominator

cos 38 cosec 52

formula

cosec a = sec (90-a)

so

cos 38 cosec 52 = cos 38 * (sec(90-52))= cos 38 * sec 38

= cos 38 * 1/ cos 38 = 1

lets solve denominator of third part

root 3 * tan 18 tan 35 tan 60 tan 72 tan 55

using formula

tan a = cot (90 - a) and

tan a = 1/ cot a

root 3 * tan 18 tan 35 tan 60 tan 72 tan 55

= root 3 * cot (90-18) cot (90- 35) tan 60 tan 72 tan 55

= root 3 * cot 72 cot 55 tan 60 tan 72 tan 55

= root 3 * (1/ tan 72) (1/ tan 55) tan 60 tan 72 tan 55

= root 3 * tan 60 = root 3 * root 3 = 3

combining all part we get

1+1 - 1/3 = 2-1/3 = 5/3 answer