Question

$\sf \red{ Find \: the \: value \: of \: :- }$
secθ(1+sinθ/cosθ+cosθ/1+sinθ)-2tan²θ
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Answer 1

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

Given expression is

$\rm :\longmapsto\:sec\theta \bigg( \dfrac{1 + sin\theta }{cos\theta } + \dfrac{cos\theta }{1 + sin\theta } \bigg) - 2 \: {tan}^{2} \theta$

$\rm \: = \: \:sec\theta \bigg(\dfrac{ {(1 + sin\theta )}^{2} + {cos}^{2}\theta }{cos\theta (1 + sin\theta )} \bigg) - 2 {tan}^{2}\theta$

We know,

$\boxed{ \bf{ \: {(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy}}$

So using this identity, we get

$\rm \: = \: \:sec\theta \bigg(\dfrac{1 + {sin}^{2} \theta + 2sin\theta + {cos}^{2}\theta }{cos\theta (1 + sin\theta )} \bigg) - 2 {tan}^{2}\theta$

$\rm \: = \: \:sec\theta \bigg(\dfrac{1 + {sin}^{2} \theta + {cos}^{2}\theta + 2sin\theta }{cos\theta (1 + sin\theta )} \bigg) - 2 {tan}^{2}\theta$

We know that,

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

So, using this, we have

$\rm \: = \: \:sec\theta \bigg(\dfrac{1 + 1 + 2sin\theta }{cos\theta (1 + sin\theta )} \bigg) - 2 {tan}^{2}\theta$

$\rm \: = \: \:sec\theta \bigg(\dfrac{2 + 2sin\theta }{cos\theta (1 + sin\theta )} \bigg) - 2 {tan}^{2}\theta$

$\rm \: = \: \:sec\theta \bigg(\dfrac{2(1 + sin\theta) }{cos\theta (1 + sin\theta )} \bigg) - 2 {tan}^{2}\theta$

$\rm \: = \: \:sec\theta \bigg(\dfrac{2 }{cos\theta } \bigg) - 2 {tan}^{2}\theta$

$\rm \: = \: \:sec\theta (2sec\theta ) - 2 {tan}^{2} \theta$

$\rm \: = \: \: {2sec}^{2}\theta - {2tan}^{2} \theta$

$\rm \: = \: \:2( {sec}^{2}\theta - {tan}^{2} \theta )$

We know,

$\boxed{ \bf{ \: {sec}^{2} x - {tan}^{2} x = 1}}$

So, using this, we have

$\rm \: = \: \:2 \times 1$

$\rm \: = \: \:2$

Hence,

The value of

$\boxed{ \bf{ \: \:sec\theta \bigg( \dfrac{1 + sin\theta }{cos\theta } + \dfrac{cos\theta }{1 + sin\theta } \bigg) - 2 \: {tan}^{2} \theta \: = \: 2}}$

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