TS ICET 24 May 2018 Shift 1 Paper

Question : 119 of 200

Marks: +1, -0

sin 0° + cos 30° — tan 45° + cosec 60° + cot 90° =

Solution:

\GE=0+\;{√{3}}/{2}-1+\;{2}/{√{3}}+0

=\;{3-2 √{3}+4}/{2 √{3}}=\;{7-2 √{3}}/{2 √{3}}

=\;{7 √{3}-6}/{6}

(multiplying and dividing with

√3

)

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