Question

if tan theta + cot theta is equal to 4 then find tan square theta + cot square theta​

Answer 1

$\huge \green{\mathfrak{solution}}$

Given:

$\boxed{tan \theta + cot \theta = 4}$

To find:

$\boxed{ {tan}^{2} \theta \: + {cot}^{2} \theta}$

$\implies \: tan \theta \: + \: cot \theta = 4$

Square both side , we get

$\implies \:( {tan \theta \: + cot \theta})^{2} = {4}^{2} \\ \\ \implies {tan}^{2} \theta + {cot}^{2} \theta \: + 2tan \theta \: cot \theta = 16 \\ \\ \implies \: {tan}^{2} \theta \: + {cot}^{2} \theta + 2 = 16 \\ \\ \implies \: {tan}^{2} \theta \: + {cot}^{2} \theta = 16 - 2 = 14 \\ \\ \implies \: {tan}^{2} \theta \: + {cot}^{2} \theta = 14$

Hence , the value is

$\huge \boxed {\red{\bold{14}}}$

◆tan@×cot@=1

Answer 2

$\large\red{\sf\underline{\underline{\:Given:\:\:}}}$

Tan@ + cot@ = 4

$\underline{{\colorbox{yellow}{Question}}}$

tan²@ + cot²@ = ?

we know that,

$\large\red{\boxed{\sf </strong><strong>(x + y)^{2} = {x}^{2} + {y}^{2} + 2xy</strong><strong>}}$

$\large\</strong><strong>g</strong><strong>r</strong><strong>e</strong><strong>e</strong><strong>n</strong><strong>{\boxed{\sf </strong><strong>\cot(x) = \frac{1}{ \tan(x) }</strong><strong>}}$

So,

Tan@ + cot@ = 4

Squaring both sides we get,

(Tan@ + cot@) = (4)²

→ Tan²@ + cot²@ + 2tan@*cot@ = 16

→ Tan²@ + cot²@ + 2tan@ × [1/tan@]= 16

→ Tan²@ + cot²@ + 2 = 16

→ $\large\</strong><strong>p</strong><strong>i</strong><strong>n</strong><strong>k</strong><strong>{\boxed{\sf </strong><strong>\tan^{2} </strong><strong>@</strong><strong> + \cot{}^{2} </strong><strong>@</strong><strong> = 14</strong><strong>}}$

$\color {red}\large\bold\star\underline\mathcal{Extra\:Brainly\:Knowledge:-}$

$\boxed{\begin{minipage}{7 cm}Fundamental Trignometric Identities\\ \\ \sin^2\theta+\cos^2\theta=1\\ \\ 1+\tan^2\theta=\sec^2\theta \\ \\ 1+\cot^2\theta=\text{cosec}^2 \, \theta \end{minipage}}$