Question

If tanA +cotA= 2,then find the value of tan square A+cot square A

Answer 1

tan A + cot A = 2
( tan A + cot A )^2 = 2^2
tan^2 A + cot^2 A + 2 tan A . cot A = 4
tan^2 A + cot^2 A + 2 = 4
tan^2 A + cot^2 A = 4 - 2
tan^2 A + cot^ A = 2

Answer 2

Step-by-step explanation:

[ Let a = ∅ ]

▶ Answer :-

→ tan²∅ + cot²∅ = 2 .

▶ Step-by-step explanation :-

➡ Given :-

→ tan ∅ + cot ∅ = 2 .

➡ To find :-

→ tan²∅ + cot²∅ .

$\huge \pink{ \mid \underline{ \overline{ \sf Solution :- }} \mid}$

We have ,

$\begin{lgathered}\begin{lgathered}\sf \because \tan \theta + \cot \theta = 2. \\ \\ \sf \implies \tan \theta + \frac{1}{ \tan \theta} = 2. \\ \\ \sf \implies \frac{ { \tan}^{2} \theta + 1}{ \tan \theta} = 2. \\ \\ \sf \implies { \tan}^{2} \theta + 1 = 2 \tan \theta. \\ \\ \sf \implies { \tan}^{2} \theta - 2 \tan \theta + 1 = 0. \\ \\ \sf \implies {( \tan \theta - 1)}^{2} = 0. \\ \\ \bigg( \sf \because {(a - b)}^{2} = {a}^{2} - 2ab + {b}^{2} . \bigg) \\ \\ \sf \implies \tan \theta - 1 = \sqrt{0} . \\ \\ \sf \implies \tan \theta - 1 = 0. \\ \\ \: \: \: \: \large \green{\sf \therefore \tan \theta = 1.}\end{lgathered}\end{lgathered}$

▶ Now,

→ To find :-

$\begin{lgathered}\begin{lgathered}\sf \because { \tan}^{2} \theta + { \cot}^{2} \theta . \\ \\ \sf = { \tan}^{2} \theta + \frac{1}{ { \tan}^{2} \theta } . \\ \\ \sf = {1}^{2} + \frac{1}{ {1}^{2} } . \: \: \: \: \bigg( \green{\because \tan \theta = 1}. \bigg) \\ \\ \sf = 1 + 1. \\ \\ \huge \boxed{ \boxed{ \orange{ = 2.}}}\end{lgathered}\end{lgathered}$

✔✔ Hence, it is solved ✅✅.