Question

this 34 no sum say me how to do it plsssssss cbse maths​

Answer 1

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

Given that,

$\rm :\longmapsto\:cosec\theta - sin\theta = l$

$\rm :\longmapsto\:\dfrac{1}{sin\theta } - sin\theta = l$

$\rm :\longmapsto\:\dfrac{1 - {sin}^{2}\theta }{sin\theta } = l$

$\rm :\longmapsto\:\dfrac{{cos}^{2}\theta }{sin\theta } = l - - - (1)$

Also, Given that

$\rm :\longmapsto\:sec\theta - cos\theta$

$\rm :\longmapsto\:\dfrac{1}{cos\theta } - cos\theta = m$

$\rm :\longmapsto\:\dfrac{1 - {cos}^{2}\theta }{cos\theta } = m$

$\rm :\longmapsto\:\dfrac{{sin}^{2}\theta }{cos\theta } = m - - - - (2)$

Now, we have to prove that

$\boxed{ \bf{ \: {l}^{2}{m}^{2}( {l}^{2} + {m}^{2} + 3)}}$

Let we first solve

$\rm :\longmapsto\: {l}^{2} + {m}^{2} + 3$

$\rm \:  =  \:  \: \dfrac{ {cos}^{4}\theta }{ {sin}^{2} \theta } + \dfrac{ {sin}^{4} \theta }{ {cos}^{2} \theta } + 3$

$\rm \:  =  \:  \: \dfrac{ {sin}^{6} \theta + {cos}^{6} \theta + 3 {sin}^{2}\theta {cos}^{2} \theta }{ {sin}^{2}\theta {cos}^{2} \theta}$

can be rewritten as

$\rm \:  =  \:  \: \dfrac{ {sin}^{6} \theta + {cos}^{6} \theta + 3 {sin}^{2}\theta {cos}^{2} \theta \times 1 }{ {sin}^{2}\theta {cos}^{2} \theta}$

$\rm \:  =  \:  \: \dfrac{ {sin}^{6} \theta + {cos}^{6} \theta + 3 {sin}^{2}\theta {cos}^{2} \theta \times ( {sin}^{2}\theta + {cos}^{2}\theta}{ {sin}^{2}\theta {cos}^{2} \theta}$

$\rm \:  =  \:  \: \dfrac{ {sin}^{6} \theta + {cos}^{6} \theta + 3 {sin}^{2}\theta {cos}^{2} \theta ( {sin}^{2}\theta + {cos}^{2}\theta}{ {sin}^{2}\theta {cos}^{2} \theta}$

$\rm \:  =  \:  \: \dfrac{{( {sin}^{2}\theta + {cos}^{2} \theta ) }^{3}}{ {sin}^{2}\theta {cos}^{2}\theta }$

$\rm \:  =  \:  \: \dfrac{ {(1)}^{3} }{ {sin}^{2}\theta {cos}^{2}\theta }$

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

Hence,

$\bf :\longmapsto\: {l}^{2} + {m}^{2} + 3 = \dfrac{1}{ {sin}^{2}\theta {cos}^{2} \theta }$

Now, Consider

$\rm :\longmapsto\: {l}^{2} {m}^{2}( {l}^{2} + {m}^{2} + 3)$

$\rm \:  =  \:  \: \dfrac{ {cos}^{4} \theta }{ {sin}^{2} \theta } \times \dfrac{ {sin}^{4} \theta }{ {cos}^{2} \theta } \times \dfrac{1}{ {sin}^{2}\theta \: {cos}^{2}\theta }$

$\rm \:  =  \:  \: 1$

Hence,

$\bf :\longmapsto\: {l}^{2} {m}^{2}( {l}^{2} + {m}^{2} + 3) = 1$

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