Question

( 1 + tanA + secA ) ( 1 + cotA - cosecA ) =​

Answer 1

$\large\underline{\sf{Solution-}}$

Given Trigonometric expression is

$\rm \: (1 + tanA + secA)(1 + cotA - cosecA)$

We know,

$\boxed{\tt{ tanx = \frac{sinx}{cosx}}} \\ \\ \boxed{\tt{ cotx = \frac{cosx}{sinx}}} \\ \\ \boxed{\tt{ secx = \frac{1}{cosx}}} \\ \\ \boxed{\tt{ cosecx = \frac{1}{sinx} \: }}$

So, using these identity, we get

$\rm \: = \: \bigg(1 + \dfrac{sinA}{cosA} + \dfrac{1}{cosA} \bigg)\bigg(1 + \dfrac{cosA}{sinA} - \dfrac{1}{sinA} \bigg)$

$\rm \: = \: \bigg(\dfrac{cosA + sinA + 1}{cosA} \bigg)\bigg(\dfrac{sinA + cosA - 1}{sinA}\bigg)$

can be further rewritten as

$\rm \: = \: \bigg(\dfrac{cosA + sinA + 1}{cosA} \bigg)\bigg(\dfrac{cosA + sinA - 1}{sinA}\bigg)$

We know

$\boxed{\tt{ (x + y)(x - y) = {x}^{2} - {y}^{2} \: }}$

So, using this identity, we get

$\rm \: = \: \dfrac{ {(cosA + sinA)}^{2} - {1}^{2} }{cosA \: sinA}$

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

We know,

$\boxed{\tt{ {sin}^{2}x + {cos}^{2}x = 1 \: }}$

So, using this identity, we get

$\rm \: = \: \dfrac{ 1 + 2sinA \: cosA - 1}{cosA \: sinA}$

$\rm \: = \: \dfrac{2sinA \: cosA}{cosA \: sinA}$

$\rm \: = \: 2$

Hence,

$\boxed{\tt{ \rm \: (1 + tanA + secA)(1 + cotA - cosecA) = 2}}$

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Additional Information:-

Relationship between sides and T ratios

sin θ = Opposite Side/Hypotenuse

cos θ = Adjacent Side/Hypotenuse

tan θ = Opposite Side/Adjacent Side

sec θ = Hypotenuse/Adjacent Side

cosec θ = Hypotenuse/Opposite Side

cot θ = Adjacent Side/Opposite Side

Reciprocal Identities

cosec θ = 1/sin θ

sec θ = 1/cos θ

cot θ = 1/tan θ

sin θ = 1/cosec θ

cos θ = 1/sec θ

tan θ = 1/cot θ

Co-function Identities

sin (90°−x) = cos x

cos (90°−x) = sin x

tan (90°−x) = cot x

cot (90°−x) = tan x

sec (90°−x) = cosec x

cosec (90°−x) = sec x

Fundamental Trigonometric Identities

sin²θ + cos²θ = 1

sec²θ - tan²θ = 1

cosec²θ - cot²θ = 1