Question

tanA+cotA =2 find value of tan^2A+cot^2A

Answer 1

Answer:

Step-by-step explanation:

tan A + cot A=2(given)

tan A square + cot A square  + 2=4  (squaring on both sides of the equation and as you know tan A * cot A=1)

therefore,  tan A square + cot A square=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 ✅✅.