Previous Year Papers

Download Solved Question Papers Free for Offline Practice and view Solutions Online.

Test Series

Take Zigya Full and Sectional Test Series. Time it out for real assessment and get your results instantly.

Test Yourself

Practice and master your preparation for a specific topic or chapter. Check you scores at the end of the test.

Introduction to Trigonometry

Short Answer Type

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

89 Views

382.

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

86 Views

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 $cos squared straight A space equals space fraction numerator straight m squared minus 1 over denominator straight n squared minus 1 end fraction.$

space space space space space space space space space space space space space tanA space equals space straight n space tan space straight B rightwards double arrow space space space space space space space space tan space straight B space equals space 1 over straight n tanA rightwards double arrow space space space space space space space space cot space straight B space equals space fraction numerator straight n over denominator tan space straight A end fraction

and space space space space sin space straight A space equals space straight m space sinB rightwards double arrow space space space space space space sin space straight B space equals space 1 over straight m sinA rightwards double arrow space space space space space space cosecB space equals space fraction numerator straight m over denominator sin space straight A end fraction

Substituting the values of cot B and cosec B in cosec² B - cot² B = 1, we get

rightwards double arrow space space space space space space space space space fraction numerator straight m squared over denominator sin squared straight A end fraction minus fraction numerator straight n squared over denominator tan squared straight A end fraction equals 1 rightwards double arrow space space space space space fraction numerator straight m squared over denominator sin squared straight A end fraction minus fraction numerator straight n squared space cos squared straight A over denominator sin squared straight A end fraction equals 1 rightwards double arrow space space space space space space space fraction numerator straight m squared minus straight n squared space cos squared straight A over denominator sin squared straight A end fraction equals 1 rightwards double arrow space space space space space space space space space space space straight m squared minus straight n squared cos squared straight A space equals space sin squared straight A rightwards double arrow space space space space space space space space space space space space straight m squared minus straight n squared cos squared straight A space equals space 1 minus cos squared straight A rightwards double arrow space space space space space space space space space space space space straight m squared minus 1 space equals space straight n squared cos squared straight A space minus space cos squared straight A rightwards double arrow space space space space space space space space space space space space fraction numerator straight m squared minus 1 over denominator straight n squared minus 1 end fraction equals cos squared straight A

143 Views

384.

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

120 Views

385.

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

90 Views

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

89 Views

388.

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

93 Views

389.

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

136 Views

390.

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

93 Views

Previous Year Papers

Test Series

Test Yourself

Short Answer Type

Long Answer Type

383. If then show that

Short Answer Type