prove cos'{cosa+cosb/1+cosacosb}=2tan'{tana/2 tanb/2}

Question

' represnt inverse

Answer 1

Question: prove cos(-1){(cosa+cosb)/(1+cosacosb)}=2tan^(-1){tana/2*tanb/2}

Let
cos^(-1)((cosa+cosb)/(1+cosa .cosb ))=s
Then
(cosa+cosb)/(1+cosa.cosb )=coss=(√(1+tan^2s ))^(-1)

Let
2 tan^(-1)(tan〖a/2〗. tan〖b/2〗)=t
Then
tan〖a/2〗. tan〖b/2〗= tan〖t/2〗

Using tan〖a/2〗. tan〖b/2〗= tan〖t/2〗 and the identity tan2A=(2 tanA)/(1-tan^2A ), then

1+ tan^2t=1+((2 tan〖t/2〗)/(1 - tan^2〖t/2〗))^2
=1+(4 tan^2〖t/2〗)/(1 - 2 tan^2〖t/2〗+ tan^4〖t/2〗)
=(1+2 tan^2〖t/2〗+ tan^4〖t/2〗)/(1 - 2 tan^2〖t/2〗+ tan^4〖t/2〗)
=(1+tan^2〖t/2〗)^2/(1 - tan^2〖t/2〗)^2

√(1+tan^2t)=(1+tan^2〖t/2〗)/(1 - tan^2〖t/2〗)
(√(1+tan^2t))^(-1)=(1 - tan^2〖t/2〗)/(1 + tan^2〖t/2〗)
(√(1+tan^2t))^(-1)=(1 - tan^2〖a/2〗.tan^2〖b/2〗)/(1 + tan^2〖a/2〗. tan^2⁡〖b/2〗)
(√(1+tan^2t ))^(-1)=(cos^2〖a/2〗. cos^2〖b/2〗- sin^2〖a/2〗. sin^2〖b/2〗)/(cos^2〖a/2〗. cos^2〖b/2〗+ sin^2〖a/2〗. sin^2〖b/2〗)

Using the identities

cos^2A=1/2 (1+cos2A )
sin^2A=1/2 (1-cos2A ),

(√(1+tan^2t ))^(-1)=((1+cosa )(1+cosb ) - (1-cosa )(1-cosb))/((1+cosa )(1+cosb ) + (1-cosa )(1-cosb))
=(1 + cosa + cosb + cosa.cosb - 1 + cosa + cosb - cosa.cosb)/(1 + cosa + cosb + cosa.cosb + 1 - cosa - cosb + cosa.cosb )
=2(cosa + cosb)/(2 + 2.cosa.cosb )
(√(1+tan^2t ))^(-1)=(cosa + cosb)/(1 + cosa.cosb )
⇒ (s=t)

Hence,

cos^(-1)((cosa + cosb)/(1 + cosa.cosb )) = 2 tan^(-1)(tan〖a/2〗. tan〖b/2〗)

Comments

is their any other method to do this question in easy manner

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Yes. See my solution.

Answer 2

Let x=cos⁻¹((cos(a)+cos(b))/(1+cos(a)cos(b)), then cos(x)=(cos(a)+cos(b))/(1+cos(a)cos(b).

sin(x)=sin(a)sin(b)/(1+cos(a)cos(b)). See below for explanation.

[sin(x)=√(1-cos²(x))=

√(1+2cos(a)cos(b)+cos²(a)cos²(b)-cos²(a)-2cos(a)cos(b)-cos²(b))/(1+cos(a)cos(b))=

√(1-cos²(a)(1-cos²(b))-cos²(b))/(1+cos(a)cos(b))=

√(1-cos²(a)sin²(b)-1+sin²(b))/(1+cos(a)cos(b))=sin(a)sin(b)/(1+cos(a)cos(b)).]

tan(x)=sin(a)sin(b)/(cos(a)+cos(b)).

Let y=2tan⁻¹(tan(a/2)tan(b/2)), so tan(y/2)=tan(a/2)tan(b/2).

But tan(a/2)≡(1-cos(a))/sin(a) [because 1-cos(a)=2sin²(a/2) and sin(a)=2sin(a/2)cos(a/2)]

and tan(b/2)≡(1-cos(b))/sin(b). These are trig identities.

Therefore tan(y/2)=(1-cos(a))(1-cos(b))/(sin(a)sin(b))=

(1+cos(a)cos(b)-cos(a)-cos(b))/(sin(a)sin(b))=

(1+cos(a)cos(b))/(sin(a)sin(b))-(cos(a)+cos(b))/(sin(a)sin(b))=

1/sin(x)-1/tan(x)=(1-cos(x))/sin(x))≡tan(x/2).

Thus tan(y/2)=tan(x/2), y/2=x/2, making x=y. Therefore,

cos⁻¹((cos(a)+cos(b))/(1+cos(a)cos(b))=2tan⁻¹(tan(a/2)tan(b/2)) QED

Comments

Alternative solution:

 

To prove:

arccos{(cos(a)+cos(b))/(1+cos(a)cos(b))}=

2arctan{tan(a/2)tan(b/2)}.

Let cos(A)=(cos(a)+cos(b))/(1+cos(a)cos(b)).

Let tan(B)=tan(a/2)tan(b/2).

So essentially we need to prove A=2B.

Let t=tan(θ/2), so we can expand the trig identity:

tan(θ)=2t/(1-t²), so t²tan(θ)+2t-tan(θ)=0.

So, t²+2tcot(θ)=1.

Completing the square:

t²+2tcot(θ)+cot²(θ)=1+cot²(θ)=csc²(θ).

(t+cot(θ))²=csc²(θ), t=±csc(θ)-cot(θ).

So tan(θ/2)=(±1-cos(θ))/sin(θ).

If we keep θ positive, tan(θ/2)=(1-cos(θ))/sin(θ).

So tan(a/2)=(1-cos(a))/sin(a) and tan(b/2)=(1-cos(b))/sin(b).

tan(a/2)tan(b/2)=((1-cos(a))/sin(a))((1-cos(b))/sin(b)).

sin(a)sin(b)=½(cos(a-b)-cos(a+b)),

cos(a)cos(b)=½(cos(a-b)+cos(a+b)).

If A=2B, then secA=sec(2B).

secA=(1+cos(a)cos(b))/(cos(a)+cos(b)).

tan(2B)=2tanB/(1-tan²B)=2tan(a/2)tan(b/2)/(1-tan²(a/2)tan²(b/2)).

2tanB=2tan(a/2)tan(b/2)=2(1-cos(a))(1-cos(b))/(sin(a)sin(b)).

tan²B=tan²(a/2)tan²(b/2)=(1-cos(a))²(1-cos(b))²/((1-cos²(a))(1-cos²(b)).

tan²B=((1-cos(a))(1-cos(b)))/((1+cos(a))(1+cos(b))).

1-tan²B=((1+cos(a))(1+cos(b))-(1-cos(a))(1-cos(b)))/((1+cos(a))(1+cos(b)))=

2(cos(a)+cos(b))/((1+cos(a))(1+cos(b))).

tan(2B)=sin(a)sin(b)/(cos(a)+cos(b)).

sec²(2B)=1+tan²(2B)=((cos(a)+cos(b))²+sin²(a)sin²(b))/(cos(a)+cos(b))².

Replace sin²(a) with 1-cos²(a) and sin²(b) with 1-cos²(b):

(cos(a)+cos(b))²+(1-cos²(a))(1-cos²(b))=

cos²(a)+2cos(a)cos(b)+cos²(b)+1-cos²(a)-cos²(b)+cos²(a)cos²(b)=(1+cos(a)cos(b))².

sec²(2B)=(1+cos(a)cos(b))²/(cos(a)+cos(b))².

sec(2B)=(1+cos(a)cos(b))/(cos(a)+cos(b))=secA. QED