Question

If 1+ sin2α = 3 sinα cosα, then values of cot α are
(a) -1, 1 (b) 0,1 (c)1, 2 (d) -1,-1
Full explanation required

Answer 1

Correct question :-

If 1 + sin²α = 3 sin α cos α, then values of cot a are

(a) -1, 1 (b) 0, 1 (c) 1, 2 (d) -1, - 1

Answer :-

Values of cot α are 1,2 [Option.c]

Explanation :-

⇒ 1 + sin²α = 3 sinα cosα

⇒ sin²α + cos²α + sin²α = 3 sin α cos α

⇒ 2sin²α + cos²α = 3 sin α cos α --(1)

Now, we will divide both sides of equation (1) by sin²α . On dividing we get :-

⇒ 2 + cos²α/sin²α = 3 cos α/sin α

⇒ 2 + cot²α = 3 cot α

⇒ 2 + cot²α - 3 cot α = 0

⇒ cot²α - cot α - 2 cot α + 2 = 0

⇒ cot α(cot α - 1) - 2(cot α - 1) = 0

⇒ (cot α - 1)(cot α - 2) = 0

⇒ cot α = 1 ; cot α = 2

Answer 2

Appropriate Question: If 1 + sin²α = 3 sinα cosα, then values of cotα are?

      a) -1, 1

      b) 0, 1

      c) 1, 2

      d) -1, -1

Know that: sin²α  + cos²α = 1

Solution:

→ 1 + sin²α = 3 sinα cosα

~By putting the knowledge point [ sin²α  + cos²α = 1  ] mentioned above.

→ sin²α  + cos²α + sin²α = 3 sinα cosα

→ 2sin²α  + cos²α - 3 sinα cosα = 0

→ 2sin²α - 3 sinα cosα + cos²α  = 0

~ By middle term splitting

→ 2sin²α - 2 sinα cosα - sinα cosα + cos²α  = 0

~By Factorization

→ 2sinα(  sinα - cosα ) - cosα( sinα - cosα ) = 0

→ ( 2sinα - cosα )( sinα - cosα ) = 0

      ___________

~As we know that, if we multiply two numbers and the result is zero, then either of them can be zero. We can get two values by first letting the value of first bracket be zero and then another bracket's value.

For the first bracket -

→ 2sinα - cosα = 0

→ 2sinα = cosα

→ cosα/sinα = 2

• → cotα = 2

For the second bracket -

→ sinα - cosα = 0

→ sinα = cosα

→ cosα/sinα = 1

• → cotα = 1

      ___________

• Henceforth, the values of cotα are either 1,2 [ Option c ]