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.

Inverse Trigonometric Functions

Short Answer Type

71. Show that $sin to the power of negative 1 end exponent 12 over 13 plus cos to the power of negative 1 end exponent 4 over 5 plus tan to the power of negative 1 end exponent 63 over 16 equals straight pi.$

Let space sin to the power of negative 1 end exponent 12 over 13 equals straight x comma space cos to the power of negative 1 end exponent 4 over 5 equals straight y comma space space space tan to the power of negative 1 end exponent 63 over 16 equals straight z therefore space space space sin space straight x equals 12 over 13 comma space cos space straight y equals 4 over 5 comma space space tan space straight z equals 63 over 16 therefore space space space cos space straight x equals square root of 1 minus sin squared space straight x end root equals square root of 1 minus 144 over 169 end root equals square root of 25 over 169 end root equals 5 over 13 therefore space space space tan space straight x equals fraction numerator sin space straight x over denominator cos space straight x end fraction equals 12 over 13 cross times 13 over 5 equals 12 over 5 Again space sin space straight y space equals space square root of 1 minus cos squared space straight y end root equals square root of 1 minus 16 over 25 end root equals square root of 9 over 25 end root equals 3 over 5 therefore space space space tan space straight y space equals space fraction numerator sin space straight y over denominator space cos space straight y end fraction equals 3 over 5 cross times 5 over 4 equals 3 over 4 therefore space space cos space straight x space equals space 5 over 13 comma space sin space straight y equals 3 over 5 comma space tan space straight x space equals 12 over 5 space and space tan space straight y space equals space 3 over 4 Now space space space tan space left parenthesis straight x plus straight y right parenthesis equals fraction numerator tan space straight x plus tan space straight y over denominator 1 asterisk times tan space straight x space tan space straight y end fraction equals fraction numerator begin display style 12 over 5 end style plus begin display style 3 over 4 end style over denominator 1 minus begin display style 12 over 5 end style cross times begin display style 3 over 4 end style end fraction equals fraction numerator 48 plus 15 over denominator 20 minus 36 end fraction equals negative 63 over 16 therefore space space space space space space space space space space space space space tan space space left parenthesis straight x plus straight y right parenthesis space space space equals space minus space tan space straight z where space space space space space space space space tan space straight z equals 63 over 16 space space space straight i. straight e. space space space straight z equals space tan to the power of negative 1 end exponent open parentheses 63 over 16 close parentheses

i.e. tan (x + y) = tan (- z) or tan (x + y) = tan

open parentheses straight pi minus straight z close parentheses

therefore

x + y = - z or x + y =

straight pi minus straight z

Since x, y and x are positive ,

straight x plus straight y not equal to negative straight z

therefore

x + y + z =

straight pi

or space space space space space space sin to the power of negative 1 end exponent 12 over 13 plus cos to the power of negative 1 end exponent 4 over 5 plus tan to the power of negative 1 end exponent 63 over 16 equals straight pi

278 Views

72. Prove that :

tan to the power of negative 1 end exponent 63 over 16 equals sin to the power of negative 1 end exponent 5 over 13 plus cos to the power of negative 1 end exponent 3 over 5

95 Views

73. Simplify

tan to the power of negative 1 end exponent open square brackets fraction numerator straight a space cos space straight x minus straight b space sin space straight x over denominator straight b space cos space straight x plus straight a space sin space straight x end fraction close square brackets comma space if space straight a over straight b space tan space straight x greater than negative 1.

123 Views

74.

Shoe that $sin to the power of negative 1 end exponent open parentheses 2 space x square root of 1 minus x squared end root close parentheses equals 2 space sin to the power of negative 1 end exponent x comma space space minus fraction numerator 1 over denominator square root of 2 end fraction less or equal than space x space less or equal than fraction numerator 1 over denominator square root of 2 end fraction$

133 Views

75.

Shoe that $sin to the power of negative 1 end exponent open parentheses 2 space x square root of 1 minus x squared end root close parentheses equals 2 space cos to the power of negative 1 end exponent space x comma space fraction numerator 1 over denominator square root of 2 end fraction space less or equal than x less or equal than 1$

126 Views

76.

Prove $sin space left parenthesis 2 space sin to the power of negative 1 end exponent space straight x right parenthesis equals space 2 space straight x square root of 1 minus straight x squared end root$

121 Views

77.

Prove $cos to the power of negative 1 end exponent space x equals 2 space sin to the power of negative 1 end exponent space square root of fraction numerator 1 minus x over denominator 2 end fraction end root equals 2 space cos to the power of negative 1 end exponent square root of fraction numerator 1 plus x over denominator 2 end fraction end root$