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Proof the trigonometric ratios of 45°​

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Let a rotating LINE OX−→− rotates about O in the anti-clockwise sense and starting from the initial POSITION OX−→− traces out ∠AOB = 45°.

Trigonometrical Ratios of 45°

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Take a point P on OY−→− and draw \(\overline{PQ}

\) perpendicular to OX−→−.

Now, ∠OPQ = 180° - ∠POQ - ∠PQO

= 180° - 45° - 90°

= 45°.

Therefore, in the △OPQ we have, ∠QOP = ∠OPQ.

Therefore, PQ = OQ = a (say).

Now,

OP2 = OQ2 + PQ2

OP2 = a2 + a2

OP2 = 2a2

Therefore, OP¯¯¯¯¯¯¯¯ = √2 a (Since, OP¯¯¯¯¯¯¯¯ is POSITIVE)

Therefore, from the right-angled △OPQ we get,

sin 45° = PQ¯¯¯¯¯¯¯¯OP¯¯¯¯¯¯¯¯=a2√a=12√=2√2

cos 45° = OQ¯¯¯¯¯¯¯¯¯OP¯¯¯¯¯¯¯¯=a2√a=12√=2√2

And tan 45° = PQ¯¯¯¯¯¯¯¯OQ¯¯¯¯¯¯¯¯¯=aa=1.

Clearly, CSC 45° = 1sin45° = √2,

sec 45° = 1cos45° = √2

And cot 45° = 1tan45° = 1

hope it helps you


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