Question

If 1+ sin2α = 3 sinα cosα, then values of cot α are
1)-1,1
2)0,1
3)1,2
4)-1,-1

Answer 1

Answer:

3) 1,2

Step-by-step explanation:

1+ sin2α = 3 sinα cos α sin2α + cos2α + sin2α = 3 sinα cos α 2 sin2α - 3sinα cos α + cos2α = 0 (2sinα -cos α)( sinα- cosα) =0 ∴ ∴ cotα = 2 or cotα = 1

Answer 2

SOLUTION

GIVEN

$\displaystyle \sf 1 + { \sin}^{2} A = 3 \sin A \cos A$

TO CHOOSE THE CORRECT OPTION

The value of cot A are

(a) - 1 , 1

(b) 0 , 1

(c) 1 , 2

(d) - 1 , - 1

EVALUATION

Here it is given that

$\displaystyle \sf 1 + { \sin}^{2} A = 3 \sin A \cos A$

We solve it as below

$\displaystyle \sf 1 + { \sin}^{2} A = 3 \sin A \cos A$

$\displaystyle \sf \implies { \sin}^{2} A + { \cos}^{2} A+ { \sin}^{2} A = 3 \sin A \cos A$

$\displaystyle \sf \implies { \cos}^{2} A+ 2{ \sin}^{2} A - 3 \sin A \cos A = 0$D

ividing both sides by sin²A we get

$\displaystyle \sf \implies { \cot}^{2} A+ 2 - 3 \cot A = 0$

$\displaystyle \sf \implies { \cot}^{2} A - 3 \cot A + 2 = 0$

$\displaystyle \sf \implies { \cot}^{2} A - 2 \cot A - \cot A+ 2 = 0$

$\displaystyle \sf \implies \cot A (\cot A - 2) -( \cot A - 2) = 0$

$\displaystyle \sf \implies (\cot A - 2) ( \cot A - 1) = 0$

$\displaystyle \sf \implies \cot A = 2 \:, \: 1$

FINAL ANSWER

Hence the correct option is (c) 1 , 2 ━━━━━━━━━━━━━━━━

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