Question

3sin²A - 2sinA - 5. ​

Answer 1

Answer:

3sinA+5cosA=5

⇒(3sinA+5cosA)

2

=25

⇒9sin

2

A+25cos

2

A+30sinAcosA=25 ……(1)

Let (3cosA−5sinA)

2

=k

⇒9cos

2

A+25sin

2

A−30sinAcosA=k …….(2)

(1)+(2)

⇒9(sin

2

A+cos

2

A)+25(cos

2

A+sin

2

A)=25+k

⇒9+25=25+k

⇒k=9.