Question

Solve this trigonometry Question of mathematics​

Answer 1

Given Question

Evaluate the following

$\sf \:(1 + {tan}^{2}\theta)(1 + sin\theta)(1 - sin\theta)(1 + cos\theta)(1 - cos\theta)(1 + {cot}^{2}\theta)$

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

Given Trigonometric function is

$\sf \:(1 + {tan}^{2}\theta)(1 + sin\theta)(1 - sin\theta)(1 + cos\theta)(1 - cos\theta)(1 + {cot}^{2}\theta)$

We know,

$\purple{\rm :\longmapsto\:\boxed{\tt{ {sec}^{2}x - {tan}^{2}x = 1}}}$

and

$\purple{\rm :\longmapsto\:\boxed{\tt{ {cosec}^{2}x - {cot}^{2}x = 1}}}$

So, using this identity, we get

$\sf = \:{sec}^{2}\theta[(1 + sin\theta)(1 - sin\theta)][(1 + cos\theta)(1 - cos\theta)]{cosec}^{2}\theta$

We know,

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

So, using this, we get

$\sf \:  =  \: {sec}^{2}\theta (1 - {sin}^{2}\theta)(1 - {cos}^{2}\theta) {cosec}^{2}\theta$

We know,

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

So, using this, we get

$\rm \:  =  \: \dfrac{1}{ {cos}^{2}\theta } \times {cos}^{2}\theta \times {sin}^{2}\theta \times \dfrac{1}{ {sin}^{2} \theta }$

$\sf\:  =  \: 1 \times 1$

$\sf\:  =  \: 1$

Hence,

$\boxed{\tt{ \sf \:(1 + {tan}^{2}\theta)(1 + sin\theta)(1 - sin\theta)(1 + cos\theta)(1 - cos\theta)(1 + {cot}^{2}\theta) = 1}}$

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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