Question

(1 + cot A - cosec A) (1 + tan A + sec A) = 2
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Answer 1

Answer:

Refer to the attachment

I hope it helps you ✌️✌️

Answer 2

Answer:

Step-by-step explanation:

We have:

$\Rightarrow \bold{(1+cotA-cosecA)(1+tanA+secA)}$

We know:

cot A = cos A/sin A

cosec A = 1/sin A

tan A = sin A/cos A

sec A = 1/cos A

Hence:

$\Rightarrow \bold{(1+\dfrac{cosA}{sinA}-\dfrac{1}{sinA})(1+\dfrac{sinA}{cosA}+\dfrac{1}{cosA} ) }$

$\Rightarrow \bold{\dfrac{(sinA+cosA-1)}{sinA}\dfrac{(cosA+sinA+1)}{cosA} }$

$\Rightarrow \bold{\dfrac{(sinA+cosA)^2-1}{sinAcosA} }$

We know:

(a + b)(a - b) = (a + b)²

(a + b)² = a² + b² + 2ab

Since:

$\Rightarrow \bold{\dfrac{sin^2A+cos^2A+2(sinA)(cosA)-1}{sinAcosA} }$

We know:

sin²A + cos²A = 1

Hence:

$\Rightarrow \bold{\dfrac{1+2sinAcosA-1}{sinAcosA} }$

So:

sin A cos A, sin A cos A get cancelled.

+1, -1 get cancelled.

$\bold{2}$

∴ L.H.S = R.H.S