Prove that cos-1((cosa+cosb)/1+cosa.cosb) = 2tan-1(tan(a/2).tan(b/2))
Answers
Answer:
Step-by-step explanation:
Let cos−1[cosa+cosb1+cosacosb]=2tan−1[tana2tanb2]=θ
=>cosθ=cosa+cosb1+cosacosb—(1)
and tanθ2=tana2tanb2—(2)
From trigonometric identity, cos(2α)=1−tan2(α)1+tan2(α)
For ease of clarity, let us use ta=tan(a2) and tb=tan(b2)
=>cosa=1−tan2(a2)1+tan2(a2)=1−t2a1+t2a—(3)
Similarly, cosb=1−t2b1+t2b—(4)
Numerator (Nr) of Equation (1) can be written as:
From (3) and (4), Nr=cosa+cosb=1−t2a1+t2a+1−t2a1+t2a
=(1−t2a)(1+t2b)+(1−t2b)(1+t2a)(1+t2a)(1+t2b)
=1+t2b−t2a−t2at2b+1+t2a−t2b−t2at2b(1+t2a)(1+t2b)
=2−2t2at2b(1+t2a)(1+t2b)=2∗1−t2at2b(1+t2a)(1+t2b)—(5)
Denominator (Dr) of Equation (1) can be written as:
Dr=1+cosacosb=1+[1−t2a1+t2a∗1−t2b1+t2b]
=(1+t2a)(1+t2b)+(1−t2a)(1−t2b)(1+t2a)(1+t2b)
=1+t2a+t2b+t2a+t2b+1−t2b−t2a−t2at2b(1+t2a)(1+t2b)
=2+2t2at2b(1+t2a)(1+t2b)
=2(1+t2at2b)(1+t2a)(1+t2b)—(6)
From (1),(5) and (6),
cosθ=Nr÷Dr
=2∗1−t2at2b(1+t2a)(1+t2b)÷2(1+t2at2b)(1+t2a)(1+t2b)
=1−t2at2b1+t2at2b
consider ta∗tb=tc
=>cosθ=1−t2c1+t2c−−(7)
where tan(θ2)=ta∗tb=tan(a2)∗tan(b2)
=>θ2=tan−1[tan(a2)∗tan(b2)]
=>θ=2tan−1[tan(a2)∗tan(b2)]
Thus proved