Question

2tan²30sec²52sin²38/(cosec²70-tan²30) please tell me​

Answer 1

Answer:

Solution:

[ There are many errors in the question , plz check ]

It may be like this:

Given

Cosec²(90-theta)-tan²theta÷4(cos²48+cos² 42) - 2tan² 30 sec² 52 sin² 28÷cosec² 70- tan² 20?

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we are using following:

i) cosec(90-theta) = sec (theta)

ii)cos 48 = cos(90-42) = sin 42

iii)tan30° = 1/√3

iv) sec 52° = sec(90-28)

= cosec 28

v)cosec70° = cosec(90-70)

= sec 20°

vi) tan 30°= 1/√3

And

a) sec²A - tan²A = 1

b) sin²A+cos²A = 1

c) cosecAsinA = 1

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Now ,

Cosec²(90-theta)-tan²theta÷4(cos²48+cos² 42) - 2tan² 30 sec² 52 sin² 28÷cosec² 70- tan² 20

= [sec² theta - tan² theta]/[4(sin²42+cos²42)]-[2tan²30cosec²28sin²28]/[sec²20-tan²20]

= 1/(4×1) - [2×(1/√3)²×1]/1

= 1/4 - 2/3

= (3-4)/12

= -1/12

Therefore,

Cosec²(90-theta)-tan²theta÷4(cos²48+cos² 42) - 2tan² 30 sec² 52 sin² 28÷cosec² 70- tan² 20 = -1/12

••••

Step-by-step explanation:

Answer 2

Answer:

2tan²30sec²52sin²38/(cosec²70-tan²30) please tell me