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11. Find the values of each of the expressions

tan to the power of negative 1 end exponent open parentheses tan space fraction numerator 3 straight pi over denominator 4 end fraction close parentheses

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12. Using principle value, evaluate the following :

cos to the power of negative 1 end exponent open parentheses cos fraction numerator 2 straight pi over denominator 3 end fraction close parentheses plus sin to the power of negative 1 end exponent open parentheses sin fraction numerator 2 straight pi over denominator 3 end fraction close parentheses.

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13.

Find the value of each of the following :

(i) sin (sin^–1x + cos^–1.x) (ii) cos (sec^–1x + cosec^–1x). | x | ≥ 1

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14. Find the principal value of sin^–1

open parentheses fraction numerator 1 over denominator square root of 2 end fraction close parentheses.

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15. Find the principal value of cot^–1

open parentheses negative fraction numerator 1 over denominator square root of 3 end fraction close parentheses

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16. Find the principal values of the following

sin to the power of negative 1 end exponent open parentheses negative 1 half close parentheses

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17. Find the principal values of the following

cos to the power of negative 1 end exponent open parentheses fraction numerator square root of 3 over denominator 2 end fraction close parentheses

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18. Find the principal values of the following

cosec^-1(2)

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19. Find the principal values of the following

tan to the power of negative 1 end exponent left parenthesis negative square root of 3 right parenthesis

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Let equals cos to the power of negative 1 end exponent open parentheses negative 1 half close parentheses equals space straight y space where space 0 space less or equal than space straight y space less or equal than space straight pi therefore space space space space cos space straight y space equals space minus 1 half space space where space 0 space less or equal than space straight y space less or equal than space straight pi rightwards double arrow space space space space space straight y equals fraction numerator 2 straight pi over denominator 3 end fraction space space space space space space space space space space space space space space space space space space open square brackets because space space space cos fraction numerator 2 straight pi over denominator 3 end fraction equals cos space open parentheses straight pi minus straight pi over 3 close parentheses equals negative cos straight pi over 3 equals negative 1 half close square brackets

therefore

required principal value =

fraction numerator 2 straight pi over denominator 3 end fraction

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