Question

i) (sin A + cos A) (sec A + cosec A) = 2 + sec A cosecA

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Answer 1

Answer:

To prove:

(sin A + cos A)(sec A + cosec A) = 2 + secA×cosecA

Proof:

Taking L.H.S,

=) (Sin A + Cos A)(sec A + cosec A)

=) Sin A Sec A + Sin A Cosec A + Cos A Sec A + Cos A cosec A

=) Sin A (1/cos A) + Sin A (1/sin A)

+ cos A (1/cosA) + cos A(1/sin A)

=) Sin A/CosA + Sin A/SinA + Cos A/cos A + cos A/sin A

=) Tan A + 1 + 1 + Cot A

=) 2 + tan A + cot A

=) 2 + tan A + (1/tanA)

=) 2 + (tan^2 A + 1) / tan A

=) 2 + sec^2 A / tan A

=) 2 + 1/cos^2 A / sin A / cos A

=) 2 + cos A / cos^2 A sin A

=) 2 + 1/cos A sin A

=) 2 + (1/cos A)(1/sin A)

=) 2 + secA cosecA

=) L.H.S = R.H.S

Hence proved.

Answer 2

There goes the answer. Hope it helps.