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Introduction to Trigonometry

Short Answer Type

381. Prove that: sinθ (1 + tanθ) + cosθ (1 + cot θ) = secθ + cosecθ

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382.

Prove that:
$sec squared straight theta minus fraction numerator sin squared straight theta minus 2 sin to the power of 4 straight theta over denominator 2 space cos to the power of 4 straight theta space minus cos squared straight theta end fraction equals 1$

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Long Answer Type

383.

If $fraction numerator tan space straight A over denominator tan space straight B end fraction equals straight n space and space fraction numerator sin space straight A over denominator sin space straight B end fraction equals straight m comma$ then show that $cos squared straight A space equals space fraction numerator straight m squared minus 1 over denominator straight n squared minus 1 end fraction.$

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384.

Prove the following identity:
$fraction numerator 1 over denominator secθ minus tanθ end fraction minus 1 over cosθ space equals space 1 over cosθ minus fraction numerator 1 over denominator secθ plus tanθ end fraction.$

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385.

Prove the following identity:
$left parenthesis secθ minus cosecθ right parenthesis space left parenthesis 1 plus tanθ plus cotθ right parenthesis space equals space secθ. space tanθ space minus space cosecθ. space cotθ$

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Short Answer Type

386.

Prove the following identity:
$fraction numerator tanθ plus secθ minus 1 over denominator tanθ minus secθ plus 1 end fraction equals fraction numerator 1 plus sinθ over denominator cosθ end fraction$

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387.

Prove the following identity:
$fraction numerator cosA over denominator 1 minus tanA end fraction minus fraction numerator sin squared straight A over denominator cosA minus sinA end fraction equals sinA plus cosA.$

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388.
Prove the following identity:
sec A (1 - sin A) (sec A + tan A) = 1

sec A (1 - sin A) (sec A + tan A) = 1
L.H.S. = sec A (1 - sin A) (sec A + tan A)
$equals space open parentheses fraction numerator 1 minus sinA over denominator cosA end fraction close parentheses open parentheses 1 over cosA plus sinA over cosA close parentheses equals open parentheses fraction numerator 1 minus sinA over denominator cosA end fraction close parentheses space open parentheses fraction numerator 1 plus sinA over denominator cosA end fraction close parentheses equals space fraction numerator left parenthesis 1 minus sinA right parenthesis thin space left parenthesis 1 plus sinA right parenthesis over denominator cos squared straight A end fraction equals space fraction numerator 1 minus sin squared straight A over denominator cos squared straight A end fraction space equals space fraction numerator cos squared straight A over denominator cos squared straight A end fraction equals space space 1 equals straight R. straight H. straight S.$
Hence, L.H.S. = R.H.S.

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389.

Prove the following identity:
$fraction numerator 1 over denominator secθ minus 1 end fraction plus fraction numerator 1 over denominator secθ plus 1 end fraction equals 2 space cosecθ. space cotθ.$

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390.

Prove the following identity:
$fraction numerator cosecθ plus cotθ over denominator cosecθ minus cotθ end fraction space equals space 1 plus 2 cot squared straight theta plus 2 cosecθ. space cotθ$