Question

SecA(1-SinA)(SecA+TanA)=1​

Answer 1

Step-by-step explanation:

secA(1- sinA)( secA+tanA)

= 1/cosA(1- sinA)( 1/cosA+sinA/cosA)

= (1- sinA/cosA)(1+sinA/cosA)

= 1-sinA × 1+ sinA/cos^2A

= (1- sinA)^2/cos^2A

= cos^2A/cos^2A

= 1

Hope it helps

Answer 2

To prove:-

$\rm sec \ A ( 1 - sin \ A)(sec \ A + tan \ A) = 1$

Solution:-

LHS

$\rm sec \ A ( 1 - sin \ A)(sec \ A + tan \ A)$

• sec A = 1/cos A

• tan A = sin A / cos A

Hence,

$\rm \dashrightarrow sec \ A ( 1 - sin \ A)\Bigg (\dfrac{1}{cos \ A} + \dfrac{sin \ A}{cos \ A}\Bigg )$

$\rm \dashrightarrow sec \ A ( 1 - sin \ A)\Bigg ( \dfrac{1+sin \ A}{cos \ A}\Bigg )$

$\rm \dashrightarrow sec \ A \Bigg ( \dfrac{( 1 - sin \ A)(1+sin \ A)}{cos \ A}\Bigg )$

• a² - b² = (a - b)(a + b)

$\rm \dashrightarrow sec \ A \Bigg ( \dfrac{ 1 - sin^2 \ A}{cos \ A} \Bigg )$

• 1 - sin² A = cos² A

$\rm \dashrightarrow sec \ A \Bigg ( \dfrac{ cos^2 \ A}{cos \ A}\Bigg )$

$\rm \dashrightarrow sec \ A (cos \ A)$

• sec A = 1÷ cos A

$\rm \dashrightarrow \dfrac{1}{cos \ A }(cos \ A)$

$\bf \dashrightarrow 1 = RHS$

Hence, Proved.