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Relation between Trigonometric Ratios

If 1+sin2 θ =...

Question

Question 4
If $1 + s i n^{2} θ = 3 s i n θ . c o s θ$ , then prove that $t a n θ = 1$ or $\frac{1}{2}$ .

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Solution

Given, $1 + s i n^{2} θ = 3 s i n θ . c o s θ$

On dividing by $s i n^{2} θ$ on both sides, we get;

$\frac{1}{s i n^{2} θ} + 1 = 3. c o t θ$ $[∵ c o t θ = \frac{c o s θ}{s i n θ}]$

$\Rightarrow c o s e c^{2} θ + 1 = 3. c o t θ$ $[c o s e c θ = \frac{1}{s i n θ}]$

$\Rightarrow 1 + c o t^{2} θ + 1 = 3. c o t θ$ $∵ c o s e c^{2} θ - c o t^{2} θ = 1$

$\Rightarrow c o t^{2} θ - 3 c o t θ + 2 = 0$

$\Rightarrow c o t^{2} θ - 2 c o t θ - c o t θ + 2 = 0$ [by splitting the middle term]

$\Rightarrow c o t θ (c o t θ - 2) - 1 (c o t θ - 2) = 0$

$\Rightarrow (c o t θ - 2) (c o t θ - 1) = 0 \Rightarrow c o t θ = 1 o r 2$

$\Rightarrow t a n θ = 1 o r \frac{1}{2}$ $[∵ t a n θ = \frac{1}{c o t θ}]$

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m

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Relation between Trigonometric Ratios

Standard X Mathematics

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