Question

2. Prove that : tan 10% tan 50º tan 70º = tan 30°​

Answer 1

Answer:

Q:tan(10)−tan(50)+tan(70)=[tan(70)−tan(50)]+tan(10)tan(10)−tan(50)+tan(70)=[tan(70)−tan(50)]+tan(10)

split the question into two parts and then proceed.

Tan(70)−tan(50)Tan(70)−tan(50)

Result of 1+tan(10)tan(10)

We need to know the identities:

Tan(A+B)=tan(A)+tan(B)1−tan(A)tan(B)Tan(A+B)=tan(A)+tan(B)1−tan(A)tan(B)

Tan(A−B)=tan(A)−tan(B)1+tan(A)tan(B)Tan(A−B)=tan(A)−tan(B)1+tan(A)tan(B)

Step 1:

Tan(70)−tan(50)=tan(60+10)−tan(60−10)=[tan(60)+tan(10)1−tan(60)∗tan(10)]−[tan(60)−tan(10)1+tan(60)∗tan(10)]Tan(70)−tan(50)=tan(60+10)−tan(60−10)=[tan(60)+tan(10)1−tan(60)∗tan(10)]−[tan(60)−tan(10)1+tan(60)∗tan(10)]

Take LCM

Tan(70)−tan(50)=(tan(60)+tan(10))∗(1+tan(60)∗tan(10))−(tan(60)−tan(10))∗(1−tan(60)∗tan(10))12−(tan(60)∗tan(10))2=(√3+tan(10))∗(1+√3∗tan(10))−(√3−tan(10))∗(1−√3∗tan(10))12−(√3∗tan(10))2=(√3+3tan(10)+tan(10)+√3tan2(10)−√3+3tan(10)+tan(10)−√3tan(10)12−(√3∗tan(10))2=8∗tan(10)12−3∗tan2(10)Tan(70)−tan(50)=(tan(60)+tan(10))∗(1+tan(60)∗tan(10))−(tan(60)−tan(10))∗(1−tan(60)∗tan(10))12−(tan(60)∗tan(10))2=(√3+tan(10))∗(1+√3∗tan(10))−(√3−tan(10))∗(1−√3∗tan(10))12−(√3∗tan(10))2=(√3+3tan(10)+tan(10)+√3tan2(10)−√3+3tan(10)+tan(10)−√3tan(10)12−(√3∗tan(10))2=8∗tan(10)12−3∗tan2(10)

Step 2:

[tan(70)−tan(50)]+tan(10)=[8∗tan(10)12−3∗tan2(10)]+tan(10)[tan(70)−tan(50)]+tan(10)=[8∗tan(10)12−3∗tan2(10)]+tan(10)

=8∗tan(10)+tan(10)−3∗tan3(10)12−3∗tan2(10)=8∗tan(10)+tan(10)−3∗tan3(10)12−3∗tan2(10)

=3∗3∗tan(10)+tan3(10)12−3∗tan2(10)=3∗3∗tan(10)+tan3(10)12−3∗tan2(10)

=3∗tan(3∗10)=3∗tan(3∗10)[we use the identitytan3A=3tana−tan3a1−3tan2a]tan3A=3tana−tan3a1−3tan2a]

Thus, we have

tan(10)−tan(50)+tan(70)=3∗tan(30)tan(10)−tan(50)+tan(70)=3∗tan(30)

=3∗1√3=3∗1√3

=√3:

HOPE IT HELPED YOU