Question

Show that tan48°tan16°tan42°tan74° =1

Answer 1

tan48 × tan42 × tan16 × tan 74 = 1

L.H.S.

= {tan48tan42} × {tan16tan74}

= {tan(90-42)tan42} × {tan(90-74)tan74}

= {cot42tan42} × {cot74tan74}

= 1×1

= 1

∴ 1 = 1

∴L.HS. = R.H.S.

Hence proved.

Answer 2

LHS

=> tan(48).tan(16).tan(42).tan(74)

=> tan48.tan16.tan(90-48).tan(90-16)

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Now, using trigonometric identity :-

=> tan(90-α) = cotα

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then,

=> tan48.tan16.cot48.cot16

=> tan48×cot48×tan16×cot16

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Again using trigonometric identity :-

$= > tan \alpha = \frac{1}{cot \alpha }$
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we get,

$= > \frac{1}{cot48} \times cot48 \times \frac{1}{cot16} \times cot16 \\ \\ = > 1 \times 1 = 1$

=> LHS
_________[PROVED]