Question

Page
Date
Prove that :-
sin (A+B) sin (A-B) 2 sin ² A- sin
A ? B -
B = cos B
cos² A
i.
ü
eos (Ato) Cos (A-BJ 2 cos²A sim ² B = cos
B = cos ²g sir RA
sin
A
in.
sin (A+B+C) 2 sin A cosB cose & cos A cos a coset cos A cosB eine
sin A sin B sine.
Cos (A + B + c)2 cos A coeB Case - cosa sin B sinensin A cosB sine-
sin A sin B coe e
tan (A + B+ () - Tan Artan B+ tan e- tan Atan B tan e
It tan Atan B-tan B tane - tane tan A​

Answer 1

Answer:

Step-by-step explanation:

To Prove :-

1 ) sin(A + B) . sin(A - B) = sin²A - sin²B

= cos²B - cos²A

2) cos(A + B).cos(A - B) = cos²A - sin²B

= cos²B - sin²A

3) sin(A + B + C) = sinAcosBcosC + cosAsinBcosC + cosAcosBsinC - sinAsinBsinC

4) cos(A + B + C) =  cosAcosBcosC - sinAsinBcosC - sinAcosBsinC - cosAsinBsinC

5) tan(A + B + C) = tanA + tanB + tanC - tanAtanBtanC/1-tanAtanB-tanBtanC-tanCtanA

Solution :-

Proving 'sin(A + B) . sin(A - B) = sin²A - sin²B  = cos²B - cos²A' :-

Taking L.H.S :-

= sin(A + B) . sin(A - B)

= (sinAcosB + cosAsinB) . (sinAcosB - cosAsinB)

= (sinAcosB)² - (cosAsinB)²

[ ∴ (a + b)(a - b) = a² - b² ]

= sin²Acos²B - cos²Asin²B

= sin²A(1 - sin²B) - (1 - sin²A)sin²B

[ ∴ sin²α + cos²α = 1]

= sin²A  - sin²Asin²B - sin²B + sin²Asin²B

= sin²A - sin²B

= (1 - cos²A) - (1 - cos²B)

[ ∴ sin²α + cos²α = 1]

= 1 - cos²A - 1 + cos²B

= cos²B - cos²A

= R.H.S

Hence Proved.

2) Proving ' cos(A + B).cos(A - B) = cos²A - sin²B = cos²B - sin²A' :-

Taking L.H.S :-

= cos(A + B) . cos(A - B)

= (cosAcosB - sinAsinB) . (cosAcosB + sinAsinB)

= (cosAcosB)^2 - (sinAsinB)^2

[ ∴ (a + b)(a - b) = a² - b² ]

= cos²Acos²B - sin²Asin²B

= cos²Acos²B - (1 - cos²A) (1 - cos²B)

= cos²Acos²B - [1 - cos²B  - cos²A + cos²Acos²B]

= cos²Acos²B - 1 + cos²B + cos²A - cos²Acos²B

= - 1 + cos²B + cos²A

[let it be equation - 1]

= -sin²B + cos²A

= cos²A - sin²B

From 1 :-

= - 1 + cos²B + cos²A

= - 1 + cos²A + cos²B

= -sin²A + cos²B

= cos²B - sin²A

= R.H.S

Hence Proved.

3) Proving ' sin(A + B + C) =sinAcosBcosC + cosAsinBcosC + cosAcosBsinC - sinAsinBsinC ' :-

Taking L.H.S :-

sin(A + B + C)

= sin((A + B) + C)

= sin(A + B).cosC + cos(A + B).sinC

= (sinAcosB + cosAsinB)cosC + (cosAcosB - sinAsinB).sinC

= sinAcosBcosC + cosAsinBcosC + cosAcosBsinC - sinAsinBsinC

= R.H.S

Hence Proved

Proving 'cos(A + B + C) = cosAcosBcosC - sinAsinBcosC - sinAcosBsinC - cosAsinBsinC ' :-

Taking L.H.S :-

cos(A + B + C)

= cos((A + B) + C)

=  cos(A + B).cosC - sin(A + B).sinC

= (cosAcosB - sinAsinB)cosC - (sinAcosB + cosAsinB).sinC

= cosAcosBcosC - sinAsinBcosC - sinAcosBsinC - cosAsinBsinC

= R.H.S

Hence Proved

Proving 'tan(A + B + C) =tanA + tanB + tanC - tanAtanBtanC/1-tanAtanB-tanBtanC-tanCtanA ' :-

Taking L.H.S :-

tan(A + B + C)

= tan((A + B) + C)

= tan(A + B) + tanC/1 - tan(A + B).tanC

$=\dfrac{\dfrac{tanA+tanB}{1-tanAtanB}+tanC}{1-\dfrac{tanA+tanB}{1-tanAtanB}tanC}$

$=\dfrac{tanA+tanB+tanC-tanAtanBtanC}{1-tanAtanB-tanAtanC-tanBtanC}$

= R.H.S

Hence Proved.