Question

if tan theta -4/tan theta =3 then sin^2 theta is​

Answer 1

Answer:

Step-by-step explanation:

We have,

$\tt{tan(\theta)-\dfrac{4}{tan(\theta)}=3}$

$\tt{\implies\,\dfrac{tan^2(\theta)-4}{tan(\theta)}=3}$

$\tt{\implies\,tan^2(\theta)-4=3\,tan(\theta)}$

$\tt{\implies\,tan^2(\theta)-3\,tan(\theta)-4=0}$

$\tt{\implies\,tan^2(\theta)-4\,tan(\theta)+tan(\theta)-4=0}$

$\tt{\implies\,tan(\theta)(tan(\theta)-4)+1(tan(\theta)-4)=0}$

$\tt{\implies\,(tan(\theta)+1)(tan(\theta)-4)=0}$

$\tt{\implies\,tan(\theta)+1=0\,\,\,\,or\,\,\,\,tan(\theta)-4=0}$

$\tt{\implies\,tan(\theta)=-1\,\,\,\,or\,\,\,\,tan(\theta)=4}$

$\tt{\implies\,sec(\theta)=\sqrt{1+(-1)^2}\,\,\,\,or\,\,\,\,sec(\theta)=\sqrt{1+(4)^2}}$

$\tt{\implies\,sec(\theta)=\sqrt{2}\,\,\,\,or\,\,\,\,sec(\theta)=\sqrt{17}}$

So,

$\tt{\implies\,sin(\theta)=-\dfrac{1}{\sqrt{2}}\,\,\,\,or\,\,\,\,sin(\theta)=\dfrac{4}{\sqrt{17}}}$

Answer 2

$\blue{\large\underline{\sf{Given- }}}$

$\rm :\longmapsto\:tan\theta - \dfrac{4}{tan\theta } = 3$

$\purple{\large\underline{\sf{To\:Find - }}}$

$\rm :\longmapsto\: {sin}^{2}\theta$

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

Given that,

$\rm :\longmapsto\:tan\theta - \dfrac{4}{tan\theta } = 3$

We know,

$\rm :\longmapsto\:\boxed{\tt{ tanx = \frac{1}{cotx}}}$

So, above can be rewritten as

$\rm :\longmapsto\:\dfrac{1}{cot\theta } - 4cot\theta = 3$

$\rm :\longmapsto\:\dfrac{1 - {4cot}^{2}\theta }{cot\theta } = 3$

$\rm :\longmapsto\:1 - {4cot}^{2}\theta = 3cot\theta$

$\rm :\longmapsto\:{4cot}^{2}\theta + 3cot\theta - 1 = 0$

On splitting the middle terms, we get

$\rm :\longmapsto\:{4cot}^{2}\theta + 4cot\theta - cot\theta - 1 = 0$

$\rm :\longmapsto\:4cot\theta (cot\theta + 1) - 1(cot\theta + 1) = 0$

$\rm :\longmapsto\:(4cot\theta - 1) (cot\theta + 1) = 0$

$\rm\implies \:cot\theta = \dfrac{1}{4} \: \: or \: \: cot\theta = - 1$

Now,

We consider first

$\red{\rm\implies \:cot\theta = \dfrac{1}{4}}$

Now, Consider

$\rm :\longmapsto\: {sin}^{2}\theta$

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

$\rm \:  =  \: \dfrac{1}{1 + {cot}^{2} \theta }$

$\rm \:  =  \: \dfrac{1}{1 + {\bigg(\dfrac{1}{4} \bigg) }^{2} }$

$\rm \:  =  \: \dfrac{1}{1 + \dfrac{1}{16} }$

$\rm \:  =  \: \dfrac{16}{17}$

$\bf\implies \: \boxed{\bf{ {sin}^{2}\theta = \frac{16}{17} \: }}$

Now, When

$\red{\rm :\longmapsto\:cot\theta = - 1}$

Now, Consider

$\rm :\longmapsto\: {sin}^{2}\theta$

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

$\rm \:  =  \: \dfrac{1}{1 + {cot}^{2} \theta }$

$\rm \:  =  \: \dfrac{1}{1 + {\bigg( - 1 \bigg) }^{2} }$

$\rm \:  =  \: \dfrac{1}{1 + 1}$

$\rm \:  =  \: \dfrac{1}{2}$

$\bf\implies \: \boxed{\bf{ {sin}^{2}\theta = \frac{1}{2} \: }}$

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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