Question

If 1+ sin α = 3 sina cosa, then values of cot a are (a) -1, 1 (b) 0,1 (c)1, 2 (d) -1,-1​

Answer 1

Correct Question :- If 1+ sin²A = 3 * sin A * cos A, then values of cot A are (a) -1, 1 (b) 0,1 (c)1, 2 (d) -1,-1

Solution :-

→ 1 + sin²A = 3 * sin A * cos A

dividing both sides by cos²A,

→ (1 + sin²A)/cos²A = (3 * sin A * cos A)/cos²A

→ (1/cos²A) + (sin²A/cos²A) = 3 * (sin A / cos A)

→ sec²A + tan²A = 3 * tan A

→ (1 + tan²A) + tan²A = 3 * tan A

→ 2tan²A - 3tan A + 1 = 0

let, tan A = x ,

→ 2x² - 3x + 1 = 0

→ 2x² - 2x - x + 1 = 0

→ 2x(x - 1) - 1(x - 1) = 0

→ (x - 1)(2x - 1) = 0

→ x = 1 and (1/2) .

therefore,

→ tan A = 1 and (1/2)

hence,

→ cot A = 1 and 2 (Option C) (Ans.)

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Answer 2

SOLUTION

GIVEN

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

TO DETERMINE

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

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

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

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

Dividing 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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