UPSEE 2017 Solved Paper

Given,
sin‌a+sin‌b=

1

√2

.......(i)
cos‌a+cos‌b=

√6

2

......(ii)
On squaring both sides in Eq. (i), we get
sin2a+sin2b+2‌sin‌a‌sin‌b=

1

2

......(iii)
And on squaring both sides in Eq. (ii), we get
cos2a+cos2b+2‌cos‌a‌cos‌b =

6

4

=

3

2

.......(iv)
Now, by adding Eqs. (iii) and (iv), we get
(sin2a+sin2b+2‌sin‌a‌sin‌b) +(cos2a+cos2b+2‌cos‌a‌cos‌b) =

1

2

+

3

2

⇒(sin2a+cos2a)+(sin2b+cos2b) +2(sin‌a‌sin‌b+cos‌a‌cos‌b)=

4

2

⇒ 1+1+2‌cos(a−b)=2
∴ cos(a−b)=0 ......(iv)
On multiplying Eqs. (i) and (ii), we get
(sin‌a+sin‌b)(cos‌a+cos‌b) =

1

√2

×

√6

2

⇒sin‌a‌cos‌a+sin‌a‌cos‌b+sin‌b‌cos‌a +sin‌b‌cos‌b=

√6

2√2

⇒(sin‌a‌cos‌a+sin‌b‌cos‌b)+(sin‌a‌cos‌b+cos‌a‌sin‌b)=

√3

2

⇒

1

2

(sin‌2‌a+sin‌2‌b)+sin(a+b) =

√3

2

⇒sin(a+b)‌cos(a−b)+sin(a+b)=

√3

2

⇒0+sin(a+b)=

√3

2

[From Eq. (v), cos(a−b)=0]
⇒sin(a+b)=

√3

2