Question

Please solve this question of trignometry.​

Answer 1

Step-by-step explanation:

We have,

$\tt \: \frac{ sec ^{2} ( \theta)sin ^{2}(\theta) - cosec^{2}( \theta) + cosec^{2} ( \theta)cos^{2}( \theta) }{sec ^{2}( \theta)sin^{2}( \theta) - cosec^{2}( \theta)cos^{2}( \theta) }$

$\tt \: = \frac{ tan^{2}(\theta) - cosec^{2}( \theta) +cot^{2}( \theta) }{tan^{2}( \theta) -cot^{2}( \theta) }$

$\tt \: = \frac{ tan^{2}(\theta) -1 }{tan^{2}( \theta) -cot^{2}( \theta) }$

$\tt \: = \frac{ tan^{2}(\theta) -1 }{tan^{2}( \theta) - \dfrac{1}{tan^{2}( \theta)} }$

$\tt \: = \frac{ tan ^{2}( \theta) \{ tan^{2}(\theta) -1 \} }{tan^{4}( \theta) - 1 }$

$\tt \: = \frac{ tan ^{2}( \theta) \{ tan^{2}(\theta) -1 \} }{ \{tan^{2}( \theta) - 1 \} \{tan^{2}( \theta) + 1 \} }$

$\tt \: = \frac{ tan ^{2}( \theta) }{ tan^{2}( \theta) + 1 }$

$\tt \: = \frac{ tan ^{2}( \theta) }{ sec^{2}( \theta) }$

$\tt = sin^{2} ( \theta)$

Answer 2

$\large\underline{\sf{Given \:Question - }}$

Prove that,

$\sf \: \dfrac{ {sec}^{2}\theta {sin}^{2}\theta - {cosec}^{2}\theta + {cosec}^{2}\theta {cos}^{2}\theta}{{sec}^{2}\theta {sin}^{2}\theta- {cosec}^{2}\theta {cos}^{2}\theta} = sin ^{2} \theta$

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

Consider,

$\rm :\longmapsto\:\sf \: \dfrac{ {sec}^{2}\theta {sin}^{2}\theta - {cosec}^{2}\theta + {cosec}^{2}\theta {cos}^{2}\theta}{{sec}^{2}\theta {sin}^{2}\theta- {cosec}^{2}\theta {cos}^{2}\theta}$

can be rewritten as

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

We know,

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

So, using this identity, we get

$\sf \: = \: \dfrac{ {sec}^{2}\theta {sin}^{2}\theta - {cosec}^{2}\theta{sin}^{2}\theta}{{sec}^{2}\theta {sin}^{2}\theta- {cosec}^{2}\theta {cos}^{2}\theta}$

We know,

$\boxed{ \bf{ \:cosec\theta \times sin\theta = 1}}$

So,

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

Now, we know,

$\boxed{ \bf{ \:secx = \frac{1}{cosx}}} \: \: and \: \: \boxed{ \bf{ \:cosecx = \frac{1}{sinx}}}$

So, using these, we get

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

$\sf \:  =  \: \dfrac{\dfrac{ {sin}^{2} \theta - {cos}^{2} \theta }{ {cos}^{2} \theta } }{\dfrac{ {sin}^{4} \theta - {cos}^{4} \theta }{ {cos}^{2}\theta {sin}^{2} \theta }}$

We know,

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

So, using this, we get

$\sf \:  =  \: [ {sin}^{2}\theta - {cos}^{2}\theta ] \times \dfrac{ {sin}^{2}\theta }{( {sin}^{2}\theta - {cos}^{2}\theta)( {sin}^{2}\theta + {cos}^{2} \theta}$

We know,

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

So, we get

$\sf \:  =  \: {sin}^{2}\theta$

Hence,

$\boxed{ \bf{ \: \sf \: \dfrac{ {sec}^{2}\theta {sin}^{2}\theta - {cosec}^{2}\theta + {cosec}^{2}\theta {cos}^{2}\theta}{{sec}^{2}\theta {sin}^{2}\theta- {cosec}^{2}\theta {cos}^{2}\theta} = sin ^{2} \theta}}$

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