Question

prove:
(1 - sinA)(1 + sinA)(1 + tan Square A) = 1​

Answer 1

Answer:

(1-sinA)(1+sinA)=1²-sin²A=1-sin²A=cos²A

and 1+tan²A=sec²A

So, (1-sinA)(1+sinA)(1+tan²A)=cos²A.sec²A

=(cos²A).(1/cos²A). ( cosA.secA=1)

=1

Answer 2

To prove :-

$\sf{ \hookrightarrow(1 - \sin A)(1 + \sin A)(1 + \tan^{2}A) = 1 }$

Solution :-

» Consider LHS

$\sf{ \dashrightarrow(1 - \sin A)(1 + \sin A)(1 + \tan^{2}A) }$

» Apply formula (1-sinA) (1+sinA) = 1-sin²A

$\sf{ \dashrightarrow(1 - \sin ^{2} A)(1 + \tan^{2}A) }$

» Apply formula (1-sin²A) = cos²A

$\sf{ \dashrightarrow( \cos ^{2} A)(1 + \tan^{2}A) }$

» Apply formula (1+tan²A) = sec²A

$\sf{ \dashrightarrow( \cos ^{2} A)( \sec^{2}A) }$

» Apply formula cos²A = 1/sec²A

$\sf{ \dashrightarrow \dfrac{( \sec^{2}A)}{( \sec^{2}A)} }$

$\sf{ \dashrightarrow \cancel\dfrac{( \sec^{2}A)}{( \sec^{2}A)} }$

$\sf{ \dashrightarrow 1}$

» LHS = RHS

Hence proved