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Question
Without using trigonometric tables, prove that:
cosec 39° cos 51° + tan 21° cot 69° – sec221° = 0
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Solution
Given: cosec 39° · cos 51° + tan 21° · cot 69° – sec2 21°
To Prove: cosec 39° cos 51° + tan 21° cot 69° – sec2 21° = 0
Proof [Step-wise]:
1. Note complementary-angle identities:
cos(51°) = sin(39°) and cot(69°)
= cot(90° – 21°)
= tan 21°
2. Evaluate the first product:
`"cosec" 39^circ · cos 51^circ = (1/sin 39^circ) · cos 51^circ`
= `(1/sin 39^circ) · sin 39^circ`
= 1
3. Evaluate the second product:
tan 21° · cot 69° = tan 21° · cot(90° – 21°)
= tan 21° · tan 21°
= tan2 21°
4. Substitute into the expression:
1 + tan2 21° – sec2 21°
5. Use the Pythagorean identity 1 + tan2θ = sec2θ (with θ = 21°):
1 + tan2 21° = sec2 21°
So 1 + tan2 21° − sec2 21°
= sec2 21° − sec2 21°
= 0
cosec 39° cos 51° + tan 21° cot 69° – sec2 21° = 0.
Hence proved.
