Question

Friend if u want 100 points ,
so solve just 3 questions I.e( 6,7,8) or any of them

Answer 1

Ans.7 According to the question

2tanα=3tanβ

tanα=(3/2)tanβ

L.H.S.

= tan(α-β)

= (tanα-tanβ)/(1+tanα.tanβ)

= {(3/2)tanβ-tanβ}/{1+(3/2)tanβ.tanβ}

= {(3tanβ-2tanβ)/2}/{(2+3tan²β)/2}

= tanβ/(2+3tan²β)

= (sinβ/cosβ)/{2+3(sin²β/cos²β)}

= (sinβ/cosβ)/{(2cos²β+3sin²β)/cos²β}

= {(sinβ/cosβ)×cos²β}/{2cos²β+3(1-cos²β)} [∵ sin²β+cos²β=1]

= (sinβcosβ)/(2cos²β+3-3cos²β)

= {2sinβcosβ}/{2(3-cos²β)}

= sin2β/(6-2cos²β)                                             [∵ sin2β=2sinβcosβ]

= sin2β/(5+1-2cos²β)

= sin2β/{5-(2cos²β-1)}                                      [∵cos2β=2cos²β-1]

= sin2β/(5-cos2β)

= R.H.S. Proved.

Answer 2

Answer:

SEE STEP BY SOLUTION

Step-by-step explanation:

a/ 2(b-c) = cos^2(A/2)/ (cosC-cosB)

lhs = a / 2 (b – c )

Now use sine rule

a/ sinA = b/ sinB = c/ sinC = k

So, lhs = a / 2 (b – c )= sinA / 2(sinB -sinC)

=SinA /2* 2 cos(B+C/2) cos (B-C/2)

= sinA / 4 cos (pi/2 – A/2) cos (B-C/2)

=2 sin(A/2) cos (A/2) / 4 sin(A/2) cos (B-C/2)

= cos (A/2) / cos(B-C/2)

=cos2(A/2)/ Cos(A/2) cos(B-C/2)

= cos2(A/2)/ Cos(90 – (B+C)/2) cos(B-C/2)

=2 cos2(A/2)/ 2 sin((B+C)/2) cos(B-C/2)

= 2 cos2(A/2)/ 2 sin((B+C)/2) cos(B-C/2)

=cos2(A/2)/ (cosC-cosB)