Question

tan 10° × tan 50° +tan 50°×tan 70° +tan 70° × tan 170° = 3 ​

Answer 1

Answer:

LET

A=tanx=t

tan60=p=√3

p²=3

② LET

B=tan(x+60)

=(t+p)/(1-tp)

③ LET

C=tan(x+120)

=(t- p)/(1+tp)

④∴

(B+C)

={(1+tp)(t+p)+(1-tp)(t-p)}/(1+tp)(1-tp)

=(2t+2tp²)/{1-t²p²}

=(2t+6t)/{1–3t²}

=8t/{1–3t²}

⑤ ∴

A+(B+C)

t+{8t/(1–3t²)}

=(t-3t³+8t)/(1–3t²)

=(9t-3t³)/(1–3t²)

=3{(3t-t³)/(1–3t²)}

=3tan3x

⑥∴

tanx+tan(x+60)+tan(x+120)≡3tan3x

Plug in x=10→

∴ tan10+tan70+tan130=3tan30.

∴ tan10+tan70—tan50=3{√3/3}=√3

∴ tan10—tan50+tan70=√3

Answer 2

Answer:

LET

A=tanx=t

tan60=p=√3

p²=3

② LET

B=tan(x+60)

=(t+p)/(1-tp)

③ LET

C=tan(x+120)

=(t- p)/(1+tp)

④∴

(B+C)

={(1+tp)(t+p)+(1-tp)(t-p)}/(1+tp)(1-tp)

=(2t+2tp²)/{1-t²p²}

=(2t+6t)/{1–3t²}

=8t/{1–3t²}

⑤ ∴

A+(B+C)

t+{8t/(1–3t²)}

=(t-3t³+8t)/(1–3t²)

=(9t-3t³)/(1–3t²)

=3{(3t-t³)/(1–3t²)}

=3tan3x

⑥∴

tanx+tan(x+60)+tan(x+120)≡3tan3x

Plug in x=10→

∴ tan10+tan70+tan130=3tan30.

∴ tan10+tan70—tan50=3{√3/3}=√3

∴ tan10—tan50+tan70=√3

Step-by-step explanation: