Question

Please help me fast I need your help

Answer 1

GIVEN :–

ㅤ

• If tan x + cot x = 2 then the value of tan²x + cot²x is :

ㅤ

(a) 3

ㅤ

(b) 1

ㅤ

(c) 2

ㅤ

(d) 0

ㅤ

ANSWER :–

ㅤ

GIVEN :–

ㅤ

• tan x + cot x = 2

ㅤ

TO FIND :–

ㅤ

• tan² x + cot² x = ?

ㅤ

SOLUTION :–

ㅤ

$\bf \implies \tan(x) + \cot(x) = 2$

ㅤ

• Square on both side –

ㅤ

$\bf \implies \{ \tan(x) + \cot(x) \}^{2} = {(2)}^{2}$

ㅤ

• Using identity –

ㅤ

$\bf \implies \large { \boxed{ \bf {(a + b)}^{2} = {a}^{2} + {b}^{2} + 2ab}}$

ㅤ

• So that –

ㅤ

$\bf \implies \tan^{2} (x)+ \cot^{2} (x) + 2 \tan(x) . \cot(x) = 4$

ㅤ

$\bf \implies \tan^{2} (x)+ \cot^{2} (x) + 2 \tan(x) . \left \{ \dfrac{1}{ \tan(x) } \right \} = 4$

ㅤ

$\bf \implies \tan^{2} (x)+ \cot^{2} (x) + 2= 4$

ㅤ

$\bf \implies \tan^{2} (x)+ \cot^{2} (x) = 4 - 2$

ㅤ

$\bf \implies \tan^{2} (x)+ \cot^{2} (x) = 2$

ㅤ

$\bf \implies \large{ \boxed{ \bf \tan^{2} (x)+ \cot^{2} (x) = 2 }}$

ㅤ

• Hence , Option (c) is correct.

Answer 2

Given ,

Tan(x) + Cot(x) = 2

Squaring on both sides , we get

$\sf \mapsto { \{Tan(x) + Cot(x) \}}^{2} = {(2)}^{2} \\ \\ \sf \mapsto {Tan}^{2} (x) + {Cot}^{2} (x) + 2Tan(x) \times Cot(x) = 4 \\ \\ \sf \mapsto {Tan}^{2} (x) + {Cot}^{2} (x) +2Tan(x) \times \frac{1}{Tan(x)} = 4 \\ \\ \sf \mapsto {Tan}^{2} (x) + {Cot}^{2} (x) + 2 = 4 \\ \\ \sf \mapsto {Tan}^{2} (x) + {Cot}^{2} (x) = 2$

Therefore ,

The correct option is (iii) i.e 2

Remmember :

$\sf {(a + b)}^{2} = {(a)}^{2} + {(b)}^{2} + 2ab \\ \\ \sf Cot(x) = \frac{1}{Tan(x) }$