Question

prove that (1+cotA-cosecA) (1+tanA+secA)=2​

Answer 1

Prove That:

(1 + cotA - cosecA)(1 + tanA + secA) = 2.

Proof:

Simplifying LHS:

$= (1+\frac{cosA}{sinA} - \frac{1}{sinA})(1+\frac{sinA}{cosA} + \frac{1}{cosA})\\\\= (\frac{sinA}{sinA}+\frac{cosA}{sinA} - \frac{1}{sinA})(\frac{cosA}{cosA}+\frac{sinA}{cosA} + \frac{1}{cosA})\\\\= (\frac{sinA \: + \: cosA \: - \: 1}{sinA})(\frac{cosA \: + \: sinA \: + \: 1}{cosA})\\\\= \frac{[(sinA \: + \: cosA) \: - \: 1]}{sinA} \times \frac{[(cosA \: + \: sinA) \: + \: 1]}{cosA}\\\\= \frac{{(sinA \: + \: cosA)}^{2} \: - \: (1)^2}{sinA.cosA}$

$= \frac{sin^{2}A \: + \: cos^{2}A \: +\:2sinA.cosA}{sinA.cosA}\\\\= \frac{1 \: + \: 2sinA.cosA \: - \: 1}{sinA.cosA}\\\\= \frac{2sinA.cosA}{sinA.cosA}\\\\= 2$

Hence Proved...

Trigonometric and Algebraic Identities Used:

$1. \: cotA = \frac{cosA}{sinA}\\\\2. \: cosecA = \frac{1}{sinA}\\\\3. \: tanA = \frac{sinA}{cosA}\\\\4. \: secA = \frac{1}{cosA}\\\\5. \: sin^{2}A \: + \: cos^{2}A = 1\\\\6. \: (a+b)(a-b) = a^{2} - b^{2}\\\\7. \: (a+b)^{2} = a^{2} \: + \: b^{2} \: + \: 2ab$

Answer 2

L.H.S = ( 1+ cota-coseca) ( 1+ tana -seca)

Step-by-step explanation:

= (1+cosa/sina-1/sina) (1+sina/cosa-1/cosa)

= (sina+cosa-1/sina) (cosa+sina-1/cosa)

= ((sina+cosa) - (1)/sina) ((cosa+sina) - (1)/cosa)

= ((sina+cosa)raise to power2 - (1)raise to power 2)/sinacosa

= sin2a+cos2a+2sinacosa-1/sinacosa

= 1+2sinacosa-1/sinacosa (because sin2a+cos2a=1)

= 2sinacosa/sinacosa

=2

=R.H.S.

HOPE IT'S HELPS YOU........