Question

Answer this question

Answer 1

and I am answering the second part also please wait 1 mint for that

Answer 2

Trigonometry,

We have,

(i), (sin A + sec A)² + (cos A + cosec A)² = (1 + sec A.cosec A)

LHS = (sinA + secA)²+ ( cosA + cosecA)²

= sin²A + sec²A + 2sinAsecA + cos²A + cosec²A + 2cosA.cosecA

= sin²A + cos²A + sec²A + cosec²A + 2cotA + 2tanA

= 1 + (sin²A + cos²A)/sin²A.cos²A + 2 (cotA + tanA)

= 1 + 1/(sin²A.cos²A) + 2{(sin²A + cos²A)/(sinA + cosA)

= 1 + 2{1/(sinA.cosA)} + 1/(sin²A.cos²A)

= 1 + 2secA.cosecA + sec²A.cosec²A

= 1( + secA.cosecA)²

= RHS

Now the second one,

First let theta equat to @

(ii), (sin@ + cosec@)²+ (cos@ + sec@)²= 7 + tan²@ + cot²@

LHS = (sin@ + cosec@)²+ (cos@ + sec@)²

= sin²@ + cosec²@ + 2 + cos²@ + sec²@ + 2

= sin²@ + cos²@ + (1 + cot²@) + 1 + tan²@ + 4

= 1 + 6 + cot²@ + tan²@

= 7 + cot²@ + tan²@

= RHS

That's it
Hope it helped (+_+)