Question

if tan theta =1 then find the value of (sec theta +cosec theta) square pls help​

Answer 1

Answer:

this is the answer of that question.

Answer 2

Answer:

the answer is 16

Step-by-step explanation:

$\tan(theta) = 1 \\</p><p>then \: we \: need \: to \: find \: the \: value \: of </p><p> \\ ( {\sec(theta) + \cosec(theta) )}^{2} </p><p>$

Now squaring both sides we get,

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

we know that

$\sec(theta) = \sqrt{1 + \tan^{2} theta }$

$\cosec(theta) = \sqrt{1 + \cot^{2} theta}$

putting on the values we get,

$( {\sec(theta) + \cosec(theta) )}^{2} </p><p> = \sec^{2} theta \: + \cosec ^{2} theta + 2\sec theta \cosec theta \\ = (1 + \tan^{2} theta) + (1 + \cot^{2} theta) + 2 \sec(theta) \cosec theta \\ = 1 + 1 + 1 + 1 + 2 \sqrt{1 + \tan^{2} theta } \sqrt{1 + \cot^{2} theta} \\ = 4 + 2 \sqrt{2} \sqrt{2} \\ = 4 + 2 \times 2 \\ = 4 + 4 = 16$