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

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

straight L. straight H. straight S. space equals space fraction numerator 1 over denominator secθ minus tanθ end fraction minus 1 over cosθ space space space space space space space space space space space space space equals space open parentheses fraction numerator 1 over denominator secθ minus tanθ end fraction cross times fraction numerator secθ plus tanθ over denominator secθ plus tanθ end fraction close parentheses minus 1 over cosθ space space space space space space space space space space space space space equals open parentheses fraction numerator secθ plus tanθ over denominator sec squared straight theta minus tan squared straight theta end fraction close parentheses minus secθ space space space space space space space space space space space space space space equals space secθ plus tanθ minus secθ space equals space tanθ

straight R. straight H. straight S. equals space 1 over cosθ minus fraction numerator 1 over denominator secθ plus tanθ end fraction space space space space space space space space space space space space space equals space 1 over cosθ minus open parentheses fraction numerator 1 over denominator secθ plus tanθ end fraction cross times fraction numerator secθ minus tanθ over denominator secθ minus tanθ end fraction close parentheses space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space secθ minus open parentheses fraction numerator secθ minus tanθ over denominator sec squared straight theta minus tan squared straight theta end fraction close parentheses space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space secθ space minus left parenthesis secθ minus tanθ right parenthesis space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space secθ minus secθ plus tanθ space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space tanθ

Hence, L.H.S. = R.H.S.

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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:
$fraction numerator tanθ plus secθ minus 1 over denominator tanθ minus secθ plus 1 end fraction equals fraction numerator 1 plus sinθ over denominator cosθ end fraction$

102 Views

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

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