Question

Please answer the underlined QUESTIONS.

Answer 1

Answer:

Step-by-step explanation:

A=60,B=30

1.) sin(A+B)=sinA cosB+cosA sinB

sin(60+30)=sin60cos30+cos60sin30

1=√3/2*√3/2+1/2*1/2

√3*√3/2*2+1/2*2

3/4+1/4=4/4=1

sin(A+B)=sinAcosB+cosAsinB=>1=1

2.)  sin(A-B)=sinAcosB-cosAsinB

   sin30=sin60cos30-sin30cos60

1/2=√3/2*√3/2-1/2*1/2

1/2=3/4-1/4

1/2=1/2

5(i) and 6(ii) are same question

Answer 2

Question :

4. If A = 60° and B = 30°, verify that:

(i)sin(A+B) =sinAcosB + cosAsinB

5. If A = 60° and B = 30°, verify that: (i)sin(A-B) = sinAcosB - cosAsinB

6. If A = 60° and B = 30°, verify that:

(i)sin(A-B) = sinAcosB - cosAsinB

Answer :

★ Question no 4.(i) ★

• A = 60°
• B = 30°

Taking L.H.S ,

sin(A+B)

☆Put A = 60° and B = 30°☆

= sin(60°+30°)

= sin90°

= 1

Taking R.H.S ,

sinAcosB + cosAsinB

☆Put A = 60° and B = 30°☆

= sin60° cos30° + cos60° sin30°

$\sf{=\frac{\sqrt{3}}{2}\times\frac{\sqrt{3}}{2}+\frac{1}{2}\times\frac{1}{2}}\\ \\ \sf{=\frac{3}{4}+\frac{1}{4}}\\ \\ \sf{=\frac{4}{4}}\\ \\ \sf{=1}$

L.H.S = R.H.S [ Verified ]

________________________

★ Question no 5.(i) ★

• A = 60°
• B = 30°

Taking L.H.S ,

sin(A-B)

☆Put A = 60° and B = 30°☆

= sin(60°-30°)

= sin30°

=$\sf{\frac{1}{2}}$

Taking R.H.S,

sinAcosB - cosAsinB

☆Put A = 60° and B = 30°☆

= sin60° cos30° - cos60° sin30°

$\sf{=\frac{\sqrt{3}}{2}\times\frac{\sqrt{3}}{2}-\frac{1}{2}\times\frac{1}{2}}\\ \\ \sf{=\frac{3}{4}-\frac{1}{4}}\\ \\ \sf{=\frac{2}{4}}\\ \\ \sf{=\frac{1}{2}}$

L.H.S = R.H.S [ Verified ]

________________________

★ Question no 6.(i) ★

• A = 60°
• B = 30°

Taking L.H.S ,

sin(A-B)

☆Put A = 60° and B = 30°☆

= sin(60°-30°)

= sin30°

=$\sf{\frac{1}{2}}$

Taking R.H.S,

sinAcosB - cosAsinB

☆Put A = 60° and B = 30°☆

= sin60° cos30° - cos60° sin30°

$\sf{=\frac{\sqrt{3}}{2}\times\frac{\sqrt{3}}{2}-\frac{1}{2}\times\frac{1}{2}}\\ \\ \sf{=\frac{3}{4}-\frac{1}{4}}\\ \\ \sf{=\frac{2}{4}}\\ \\ \sf{=\frac{1}{2}}$

L.H.S = R.H.S [ Verified ]

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