Question

Tan 20 / cosec 70 whole square + cot 20 divided by sec 70 square + 2 tan 15 tan 37 tan 53 tan 60 tan 75

Answer 1

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Answer 2

Answer:

Step-by-step explanation:

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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