Question

show that
(sec A - Cos A). (cot A + tan A) = sec A . tan A​

Answer 1

Answer:

(sec A - Cos A). (cot A + tan A) = sec A . tan A

L.H.S

$( \frac{1}{cos \: a} - cos \: a)( \frac{cos \: a}{sin \: a} + \frac{sin \: a}{cos \: a} )$

$( \frac{1 - cos {}^{2} a}{cos \: a} )( \frac{cos {}^{2} a+ sin {}^{2}a }{sin \: a \: cos \: a} )$

$( \frac{sin {}^{2}a }{cos \: a} )( \frac{1}{sin \: a \: cos \: a} )$

$( \frac{sin \: a}{cos \: a} \times \frac{1}{cos \: a} )$

tan A . sec A

hence verified

Answer 2

Answer:

put cosA=1/sec A and cot A=1/ tan A

Put The Value :

(sec A - 1/sec A) . (1/ tan A+ tan A) = sec A . tan A

(sec²A-1/sec A) . ( tan² A + 1 / tan A ) = sec A. tanA

( tan²A / sec A) . ( sec²A / tan A) = sec A . tan A

tan²A/ sec A * sec²A / tan A = sec A . tan A

tan A . sec A = sec A . tan A

Hence proved!