tan68∘+sec68∘, when expressed in terms of angles between 0∘ and 45∘, becomes

We know thatsinA=cos(90˚−A)tanA=cot(90˚−A)secA=cosec (90˚−A)Using these, we gettan68∘=tan(90˚−22˚)=cot22∘sec68∘=sec (90˚−22˚)=cosec 22˚∴tan68∘+sec68∘=cot22∘+cosec 22∘.

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68^{} + 68^{}, when expressed in terms of angles between 0^{

Question

$tan 6 8^{\circ} + sec 6 8^{\circ}$ , when expressed in terms of angles between $0^{\circ}$ and $4 5^{\circ}$ , becomes

cot 2 2^{\circ} + cosec 2 2^{\circ}

sin 2 2^{\circ} + cos 2 2^{\circ}

cos 2 2^{\circ} + tan 2 2^{\circ}

None of these

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Michael Discussed

tan 6 8^{\circ} + sec 6 8^{\circ}

, when expressed in terms of angles between

0^{\circ}

and

4 5^{\circ}

, becomes

7 mins ago

Discuss this question LIVE

7 mins ago

Text solutionVerified

We know that

sin A = cos (90˚ - A)

tan A = cot (90˚ - A)

sec A = cosec (90˚ - A)

Using these, we get

tan 6 8^{\circ} = tan (90˚ - 22˚) = cot 2 2^{\circ}

sec 6 8^{\circ} = sec (90˚ - 22˚) = cosec 22˚

∴ tan 6 8^{\circ} + sec 6 8^{\circ} = cot 2 2^{\circ} + cosec 2 2^{\circ}

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Question Text	$tan 6 8^{\circ} + sec 6 8^{\circ}$ , when expressed in terms of angles between $0^{\circ}$ and $4 5^{\circ}$ , becomes
Answer Type	Text solution:1
Upvotes	150

tan68∘+sec68∘, when expressed in terms of angles between 0∘ and 45∘, becomes

Text solutionVerified

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$tan 6 8^{\circ} + sec 6 8^{\circ}$ , when expressed in terms of angles between $0^{\circ}$ and $4 5^{\circ}$ , becomes