Question

Hello.Plz help me solve it.
2/3 csc(58)^2 - 2/3 cot(58) tan32 - 5/3 tan13 tan37 tan 45 tan 53 tan 77 =1

Answer 1

[{2/3 cosec^2 (58)} - {2/3 (cot 58)( tan 32)} - {5/3 (tan 13)( tan 37 )( tan 45 )( tan 53 )( tan 77)}]
Using: tan x = cot (90 - x)

={[2/3 cosec^2 58} -{2/3 cot^2 58} - {5/3 (tan 13 tan 37 tan 45 cot 37 cot 13}]
=2/3(cosec^2 58 - cot^2 58) - 5/3 [since, tan 45 = 1]
=2/3 -5/3 [Using cosec^2 x - cot^2 x = 1]
= -1

Answer 2

Answer:

Step-by-step explanation:

PLEASE MARK ME AS BRAINLIEST

Step-by-step explanation:

It is tan 53 not tan 13.

2/3cosec^2 58 - 2/3cot58·tan32 - 5/3 tan53·tan37·tan45·tan53·tan37= -1

L.H.S

2/3cosec^2 58 - 2/3cot58·tan32 - 5/3 tan53·tan37·tan45·tan53·tan37

2/3cosec^2 58 - 2/3cot58·tan(90-32) - 5/3 tan(90-53)·tan37·tan45·tan(90-53)·tan37                                                            {By complementary angles}

2/3cosec^2 58 - 2/3cot58·cot58 - 5/3 cot37·tan37·tan45·cot37·tan37

2/3cosec^2 58 - 2/3cot^2 58 - 5/3 1/tan37·tan37·tan45·1/tan37·tan37 {cotΘ=1/tanΘ}

2/3[cosec^2 58 - cot^2 58] - 5/3 ·tan45

2/3[1] - 5/3 ·1                       {cosec^2Θ-cot^2Θ=1 ; tan45=1}

2/3-5/3

-3/3 = -1

Therefore L.H.S=R.H.S

So 2/3cosec^2 58 - 2/3cot58·tan32 - 5/3 tan53·tan37·tan45·tan53·tan37= -1