Question

if tan theta=1/root 5 find the value of cosec2theta-sec2theta/cosec2theta + sec2 theta

Answer 1

Answer:

$Value \:of\:\\</p><p>\frac{cosec^{2}\theta-sec^{2}\theta}{cosec^{2}\theta+sec^{2}\theta}=\frac{2}{3}$

Step-by-step explanation:

$Given \: tan\theta = \frac{1}{\sqrt{5}}$

$\implies tan^{2}\theta = \frac{1}{5} --(1)$

$cot^{2}\theta \\= \frac{1}{tan^{2}\theta}\\=\frac{1}{\frac{1}{5}}\\=5--(2)$

$Now,\\Value \:of\:\\</p><p>\frac{cosec^{2}\theta-sec^{2}\theta}{cosec^{2}\theta+sec^{2}\theta}\\=\frac{(1+cot^{2}\theta)-(1+tan^{2}\theta)}{(1+cot^{2}\theta)+(1+tan^{2}\theta)}$

/*By Trigonometric identities:

i) cosec²A = 1+ cot²A

ii) sec²A = 1+ tan²A */

$=\frac{1+cot^{2}\theta-1-tan^{2}\theta}{1+cot^{2}\theta+1+tan^{2}\theta}$

$=\frac{cot^{2}\theta-tan^{2}\theta}{2+cot^{2}\theta+tan^{2}\theta}$

$=\frac{5-\frac{1}{5}}{2+5+\frac{1}{5}}$

\* From (1) & (2) *\

$=\frac{\frac{(25-1)}{5}}{7+\frac{1}{5}}\\=\frac{\frac{24}{5}}{\frac{36}{5}}\\=\frac{24}{36}\\=\frac{2}{3}$

Therefore,

$Value \:of\:\\</p><p>\frac{cosec^{2}\theta-sec^{2}\theta}{cosec^{2}\theta+sec^{2}\theta}=\frac{2}{3}$

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Answer 2

Step-by-step explanation:

this answer is correct