Question

1f sin a + sin b = a and Cos a - Cos b = 5, then tan a-b/2

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Answer 1

$\bold\pink{\fbox{\sf{Solution}}}$

Explanation:

sinA+sinB=x

⇒2⋅sin(A+B2)⋅cos(A−B2)=x ...(1)

cosA+cosB=y

⇒2⋅cos(A+B2)⋅cos(A−B2)=y ...(2)

Now, we have to remove the term of sine and cosine value of  (A+B2) to get the value of tan(A−B2).

From 1st equation,

⇒4⋅sin2(A+B2)⋅cos2(A−B2)=x2

⇒4⋅cos2(A−B2)⋅{1−cos2(A+B2)}=x2 ...(3)

Similarly, from 2nd equation,

⇒4⋅cos2(A+B2)⋅cos2(A−B2)=y2

⇒4⋅cos2(A+B2)⋅cos2(A−B2)=y2 ...(4)

From 3rd and 4th equation, we get

⇒4⋅cos2(A−B2)⋅⎧⎪⎨⎪⎩1−y24⋅cos2(A−B2)⎫⎪⎬⎪⎭=x2

⇒4⋅cos2(A−B2)−y2=x2

⇒4⋅cos2(A−B2)=x2+y2

⇒cos2(A−B2)=x2+y24

⇒sec2(A−B2)=4x2+y2

⇒1+tan2(A−B2)=4x2+y2

⇒tan2(A−B2)=4x2+y2−1

⇒tan2(A−B2)=4−x2−y2x2+y2

⇒tan(A−B2)=±√4−x2−y2x2+y2

Hope it helps

Answer 2

$\huge \tt \blue {AnSWER}$

sinA+sinB=x

⇒2⋅sin(A+B2)⋅cos(A−B2)=x ...(1)

cosA+cosB=y

⇒2⋅cos(A+B2)⋅cos(A−B2)=y ...(2)

Now, we have to remove the term of sine and cosine value of (A+B2) to get the value of tan(A−B2).

From 1st equation,

⇒4⋅sin2(A+B2)⋅cos2(A−B2)=x2

⇒4⋅cos2(A−B2)⋅{1−cos2(A+B2)}=x2 ...(3)

Similarly, from 2nd equation,

⇒4⋅cos2(A+B2)⋅cos2(A−B2)=y2

⇒4⋅cos2(A+B2)⋅cos2(A−B2)=y2 ...(4)

From 3rd and 4th equation, we get

⇒4⋅cos2(A−B2)⋅⎧⎪⎨⎪⎩1−y24⋅cos2(A−B2)⎫⎪⎬⎪⎭=x2

⇒4⋅cos2(A−B2)−y2=x2

⇒4⋅cos2(A−B2)=x2+y2

⇒cos2(A−B2)=x2+y24

⇒sec2(A−B2)=4x2+y2

⇒1+tan2(A−B2)=4x2+y2

⇒tan2(A−B2)=4x2+y2−1

⇒tan2(A−B2)=4−x2−y2x2+y2

⇒tan(A−B2)=±√4−x2−y2x2+y2

Hope it helps