Question

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Answer 1

Step-by-step explanation:

QuEsTiOn:-

Prove that

tan7° × tan23° × tan60° × tan67° × tan 83° = √3

AnSwEr :-

LHS = tan7° × tan23° × tan60° × tan67° × tan 83°

we know that ,. θ

$tan ( 90 - θ ) = cotθ \: and \: cotθ = \frac{1}{tanθ}$

tan7° = tan (90° − 83° ) = cot83°

tan23° = tan (90° − 67° ) = cot67°

∴ tan7° × tan23° × tan60° × tan67° × tan83°

= cot83° × cot67° × tan60° × tan67° × tan83°

$\frac{1}{tan83°} \times \frac{1}{tan67°} \times \times tan60° \times tan67° \times tan83°$

$\frac{tan60° \times tan60° \times tan83°}{tan83° \times tan60°}$

= tan60°

= √3

∴ tan7° × tan23° × tan60° × tan67° × tan 83° = √3

Hence proved ✓

Answer 2

Answer:

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Prove that

tan7° × tan23° × tan60° × tan67° × tan 83° = √3

AnSwEr :-

LHS = tan7° × tan23° × tan60° × tan67° × tan 83°

we know that ,. θ

tan ( 90 - θ ) = cotθ \: and \: cotθ = \frac{1}{tanθ}tan(90−θ)=cotθandcotθ=

tanθ

1

tan7° = tan (90° − 83° ) = cot83°

tan23° = tan (90° − 67° ) = cot67°

∴ tan7° × tan23° × tan60° × tan67° × tan83°

= cot83° × cot67° × tan60° × tan67° × tan83°

\frac{1}{tan83°} \times \frac{1}{tan67°} \times \times tan60° \times tan67° \times tan83°

tan83°

1

×

tan67°

1

××tan60°×tan67°×tan83°

\frac{tan60° \times tan60° \times tan83°}{tan83° \times tan60°}

tan83°×tan60°

tan60°×tan60°×tan83°

= tan60°

= √3

∴ tan7° × tan23° × tan60° × tan67° × tan 83° = √3

Hence proved ✓

hope it helps u sis