Question

prove that sec A(1-sinA) (sac A+tan A) =1​

Answer 1

Answer:

Explanation:

To Prove :

• secA(1 - sinA) (secA + tanA) = 1

Formula to be used :

• secA = 1/cosA
• 1 - sin²A = cos²A
• (a - b) (a + b) = a² - b²
• tanA = sinA/cosA

Proof :

LHS

secA(1 - sinA) (secA + tanA)

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

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

=> (1 - sinA) (1 + sinA)/cosA × cosA

=> (1² - sin²A)/cos²A

=> 1 - sin²A/cos²A

=> cos²A/cos²A

=> 1

RHS

• Hence Proved!!

Answer 2

$\bf{\bold{\purple{Question ↴}}}$

Prove that sec A(1-sinA)( sec A + tan A) =1

$\bf{\bold{\purple{Forumulae~used ↴}}}$

• SecA = $\bf\dfrac{1}{cosA}$

• 1 - sin²A = cos²A

• tanA = $\bf\dfrac{sinA}{cosA}$

• (a - b) (a + b) = a² - b²

$\bf{\bold{\purple{Answer ↴}}}$

LHS

SecA(1-sinA) (secA + tana) = 1

↬$\bf\dfrac{1}{cosA}$ (1-sinA) ($\bf\dfrac{1}{cosA}$ + $\bf\dfrac{sinA}{cosA}$)

↬$\bf\dfrac{ (1-sinA}{cosA}$ ( $\bf\dfrac{1 + sinA}{cosA}$)

↬$\bf\dfrac{(1 - sinA) (1 + sinA) }{cosA × cosA}$

↬$\bf\dfrac{ (1² - sin²A) }{ cos²A}$

↬$\bf\dfrac{cos²A}{cos²A}$

↬1 = RHS

$\bf{\bold{\purple{Hence~proved}}}$