If - - -3- find the value of cosec2 - - 2 -cosec2 - - 2 -

Given : θ =√( 3 ) ⇒θ =π #160; 6 cosec π6=2, π6 = 2 √( 3 ) #160; → cosec ^ 2 θ +^ 2 θ #160; cosec ^ 2 θ ^ 2 θ #160; = 4+3 4 4 ...

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Standard XII

Mathematics

Principal Solution of Trigonometric Equation

If θ = √3, ...

Question

If $cot θ = \sqrt{3}$ , find the value of $\frac{c o s e c^{2} θ + {cot}^{2} θ}{c o s e c^{2} θ - {sec}^{2} θ}$

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Solution

Given : $cot θ = \sqrt{3}$
$\Rightarrow θ = \frac{π}{6}$

$c o s e c \frac{π}{6} = 2, sec \frac{π}{6} = \frac{2}{\sqrt{3}}$

$\to \frac{{c o s e c}^{2} θ + {cot}^{2} θ}{{c o s e c}^{2} θ - {sec}^{2} θ} = \frac{4 + 3}{4 - \frac{4}{3}} = \frac{21}{8}$

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If cotθ=√3, find the value of cosec2θ+cot2θcosec2θ−sec2θ

Given : cotθ=√3⇒θ=π6cosecπ6=2,secπ6=2√3→cosec2θ+cot2θcosec2θ−sec2θ=4+34−43=218

If $cot θ = \sqrt{3}$ , find the value of $\frac{c o s e c^{2} θ + {cot}^{2} θ}{c o s e c^{2} θ - {sec}^{2} θ}$

Given : $cot θ = \sqrt{3}$
$\Rightarrow θ = \frac{π}{6}$

$c o s e c \frac{π}{6} = 2, sec \frac{π}{6} = \frac{2}{\sqrt{3}}$

$\to \frac{{c o s e c}^{2} θ + {cot}^{2} θ}{{c o s e c}^{2} θ - {sec}^{2} θ} = \frac{4 + 3}{4 - \frac{4}{3}} = \frac{21}{8}$