Question

question number 51
see attachment

Answer 1

Answer:

Step-by-step explanation:

We have,

$\tt{\left(sec\,\theta-cosec\,\theta\right)\left(1+tan\,\theta+cot\,\theta\right)}$

$\sf{=\left(\dfrac{1}{cos\,\theta}-\dfrac{1}{sin\,\theta}\right)\left(1+\dfrac{sin\,\theta}{cos\,\theta}+\dfrac{cos\,\theta}{sin\,\theta}\right)}$

$\sf{=\left(\dfrac{sin\,\theta-cos\,\theta}{sin\,\theta\,cos\,\theta}\right)\left(\dfrac{sin\,\theta\,cos\,\theta+sin^2\,\theta+cos^2\,\theta}{sin\,\theta\,cos\,\theta}\right)}$

$\sf{=\left(\dfrac{sin\,\theta-cos\,\theta}{sin\,\theta\,cos\,\theta}\right)\left(\dfrac{1+sin\,\theta\,cos\,\theta}{sin\,\theta\,cos\,\theta}\right)}$

$\sf{=\left\{\dfrac{(sin\,\theta-cos\,\theta)(1+sin\,\theta\,cos\,\theta)}{sin^2\,\theta\cdot\,cos^2\,\theta}\right\}}$

$\sf{=\left\{\dfrac{sin\,\theta-cos\,\theta+sin^2\,\theta\,cos\,\theta-sin\,\theta\,cos^2\,\theta}{sin^2\,\theta\cdot\,cos^2\,\theta}\right\}}$

$\sf{=\left\{\dfrac{sin\,\theta-sin\,\theta\,cos^2\,\theta-cos\,\theta+sin^2\,\theta\,cos\,\theta}{sin^2\,\theta\cdot\,cos^2\,\theta}\right\}}$

$\sf{=\left\{\dfrac{sin\,\theta(1-cos^2\,\theta)-cos\,\theta(1-sin^2\,\theta)}{sin^2\,\theta\cdot\,cos^2\,\theta}\right\}}$

$\sf{=\left\{\dfrac{sin\,\theta\cdot\,sin^2\,\theta-cos\,\theta\cdot\,cos^2\,\theta}{sin^2\,\theta\cdot\,cos^2\,\theta}\right\}}$

$\sf{=\left\{\dfrac{sin^3\,\theta-cos^3\,\theta}{sin^2\,\theta\cdot\,cos^2\,\theta}\right\}}$

$\sf{=\dfrac{sin^3\,\theta}{sin^2\,\theta\cdot\,cos^2\,\theta}-\dfrac{cos^3\,\theta}{sin^2\,\theta\cdot\,cos^2\,\theta}}$

$\sf{=\dfrac{sin\,\theta}{cos^2\,\theta}-\dfrac{cos\,\theta}{sin^2\,\theta}}$

$\sf{=\dfrac{1}{cos\,\theta}\cdot\,\dfrac{sin\,\theta}{cos\,\theta}-\dfrac{1}{sin\,\theta}\cdot\dfrac{cos\,\theta}{sin\,\theta}}$

$\sf{=sec\,\theta\cdot\,tan\,\theta-cosec\,\theta\cdot\,cot\,\theta}$

Answer 2

Given Question :-

Prove that

$\sf \:(sec\theta - cosec\theta )(1 + tan\theta + cot\theta ) = sec\theta tan\theta - cosec\theta cot\theta$

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

Consider LHS

$\rm :\longmapsto\:\sf \:(sec\theta - cosec\theta )(1 + tan\theta + cot\theta )$

We know,

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

So, on substituting these Identities, we get

$\rm \:  =  \: \bigg[\dfrac{1}{cos\theta } - \dfrac{1}{sin\theta } \bigg]\bigg[1 + \dfrac{sin\theta }{cos\theta } + \dfrac{cos\theta }{sin\theta } \bigg]$

$\rm \:  =  \: \bigg[\dfrac{sin\theta - cos\theta }{sin\theta \: cos\theta } \bigg]\bigg[\dfrac{sin\theta cos\theta + {sin}^{2} \theta + {cos}^{2} \theta }{sin\theta \: cos\theta } \bigg]$

We know,

$\purple{\boxed{ \tt{ \: (x - y)( {x}^{2} + {y}^{2} + xy) = {x}^{3} - {y}^{3} \: }}}$

So, using this identity, we get

$\rm \:  =  \: \dfrac{ {sin}^{3} \theta - {cos}^{3} \theta }{ {sin}^{2}\theta \: {cos}^{2}\theta }$

$\rm \:  =  \: \dfrac{ {sin}^{3} \theta}{ {sin}^{2}\theta \: {cos}^{2}\theta } - \dfrac{ {cos}^{3} \theta }{ {sin}^{2} \theta \: {cos}^{2}\theta }$

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

$\rm \:  =  \: \dfrac{sin\theta }{ cos\theta \times cos\theta } - \dfrac{cos\theta }{ sin\theta \times sin\theta }$

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

$\rm \:  =  \: tan\theta \: sec\theta \: - \: cot\theta \: cosec\theta$

Hence,

$\boxed{ \tt{ \: \sf \:(sec\theta - cosec\theta )(1 + tan\theta + cot\theta ) = sec\theta tan\theta - cosec\theta cot\theta}}$

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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