Question

if tan theta =1. find the value of 5 cot square theta +sin square theta - 1​

Answer 1

Given :-

${tan\theta= 1}$

To find :-

${5cot^2\theta + sin^2 \theta -1}$

SOLUTION:-

1 can be written as tan45° So,

${tan\theta= tan45^{\circ}}$ So,

${\theta = 45^{\circ}}$

Substitute value of theta We get the answer

${5cot^2\theta + sin^2 \theta -1}$

${5cot^2 45^{\circ} + sin^2 45^{\circ} - 1}$

As all we know that

cot45° = 1

sin45° = 1/√2

$= 5(1) {}^{2} + \bigg(\dfrac{1}{ \sqrt{2} }\bigg) {}^{2} - 1$

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

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

$= 4+ \dfrac{1}{2}$

$= \dfrac{4\times2 +1}{2}$

$= \dfrac{8+1}{2}$

$= \dfrac{9}{2}$

So, the value of ${5cot^2\theta + sin^2 \theta -1} = \dfrac{9}{2}$

Know more :-

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\sf Trigonometry\: Table \\ \begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\boxed{\begin{array}{ |c |c|c|c|c|c|} \bf\angle A & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} &1 \\ \\ \rm cos \: A & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan A & 0 & \dfrac{1}{ \sqrt{3} }&1 & \sqrt{3} & \rm \infty \\ \\ \rm cosec A & \rm \infty & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm \infty \\ \\ \rm cot A & \rm \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0\end{array}}}\end{gathered}\end{gathered}\end{gathered} \end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}$