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Answers
Step-by-step explanation:
1+sin^2 \theta = 3sin\theta cos\ \theta \text{ then } tan\ \theta = \frac{1}{2}\ or\ 11+sin
2
θ=3sinθcos θ then tan θ=
2
1
or 1
Solution:
Given that, we have to prove:
1+sin^2 \theta = 3sin\theta cos\ \theta1+sin
2
θ=3sinθcos θ
To prove:
tan\ \theta = \frac{1}{2}\ or\ 1tan θ=
2
1
or 1
From given,
1+sin^2 \theta = 3sin\theta cos\ \theta1+sin
2
θ=3sinθcos θ
Divide\ by\ cos^2 \theta\ on\ both\ sidesDivide by cos
2
θ on both sides
\begin{gathered}\frac{1}{cos^2 \theta} + \frac{sin^2 \theta}{cos^2 \theta} = \frac{3sin\theta cos\theta}{cos^2 \theta}\\\\\frac{1}{cos^2 \theta} + \frac{sin^2 \theta}{cos^2 \theta} = \frac{3sin\theta}{cos\ \theta}\end{gathered}
cos
2
θ
1
+
cos
2
θ
sin
2
θ
=
cos
2
θ
3sinθcosθ
cos
2
θ
1
+
cos
2
θ
sin
2
θ
=
cos θ
3sinθ
We know that,
\frac{1}{cos \theta} = sec\ \theta \text{ and } \frac{sin \theta}{cos \theta} = tan \theta
cosθ
1
=sec θ and
cosθ
sinθ
=tanθ
Therefore,
\begin{gathered}sec^2 \theta + tan^2 \theta = 3tan \theta\\\\We\ know\ that\\\\1 + tan^2 \theta = sec^2 \theta\\\\Therefore\\\\1 + tan^2 \theta + tan^2 \theta = 3 tan\theta\\\\1 + 2tan^2 \theta - 3tan \theta = 0\\\\2tan^2 \theta - 3tan \theta + 1 = 0\\\\Split\ the\ middle\ term\\\\2tan^2 \theta -2tan \theta - tan\ \theta + 1 = 0\\\\\end{gathered}
sec
2
θ+tan
2
θ=3tanθ
We know that
1+tan
2
θ=sec
2
θ
Therefore
1+tan
2
θ+tan
2
θ=3tanθ
1+2tan
2
θ−3tanθ=0
2tan
2
θ−3tanθ+1=0
Split the middle term
2tan
2
θ−2tanθ−tan θ+1=0
\begin{gathered}2tan \theta (tan \theta - 1) -(tan \theta - 1) = 0\\\\(2 tan \theta - 1)(tan \theta - 1) = 0\\\\Equate\ to\ 0\\\\2 tan \theta - 1 = 0\\\\2tan \theta = 1\\\\\boxed{tan\ \theta = \frac{1}{2}}\\\\Also,\\\\tan \theta - 1 = 0\\\\\boxed{tan \theta =1}\end{gathered}
2tanθ(tanθ−1)−(tanθ−1)=0
(2tanθ−1)(tanθ−1)=0
Equate to 0
2tanθ−1=0
2tanθ=1
tan θ=
2
1
Also,
tanθ−1=0
tanθ=1
Thus proved
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