Question

2tan²30sec²52sin²38/(cosec²70-tan²30) this is the right question please help me​

Answer 1

Step-by-step explanation:

= 2×( 1/√3)^2× sec^2(90°-38°)sin^2 38

[ cosec^2( 90°-30°)tan^2 30 ]

= 2/3 × cosec^2 38 × sin^2 38

[ sec^2 30 - tan^2 30 ] [ use sqr formula: 1 + tan^2=sec^2 ]

= 2/3× 1/sin^2 38× sin^2 38. ( sin gets cancelled)

1

= 2/3× 1

= 2/3

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