Without using tables, evaluate the following:cos22o2sin68o−5tan75o2cot15o−53tan45o.tan20o.tan40o.tan50o.tan70o

We have cos22o2sin68o−5tan75o2cot15o−53tan45o.tan20o.tan40o.tan50o.tan70o=cos22o2sin(90o−22o)−5tan(90o−15o)2cot15o−53tan45o.tan20o.tan40o.tan(90o−40o).tan(90o−20o)=cos22o2cos22o−5cot15o2cot15o−53tan45o.tan20o.tan40o.cot40o.cot20o=2−52−53tan45o.(tan20o.cot20o).(tan40o.cot40o)=2−52−53.1.1.1=2−52−53=2−1=1.

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Without using tables, evaluate the following: {2 68^o }{ 22^o}

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Without using tables, evaluate the following:
$\frac{2 sin 6 8 ^{o}}{cos 2 2 ^{o}} - \frac{2 cot 1 5 ^{o}}{5 tan 7 5 ^{o}} - \frac{3 tan 4 5 ^{o} . tan 2 0 ^{o} . tan 4 0 ^{o} . tan 5 0 ^{o} . tan 7 0 ^{o}}{5}$

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Without using tables, evaluate the following:

\frac{2 sin 6 8 ^{o}}{cos 2 2 ^{o}} - \frac{2 cot 1 5 ^{o}}{5 tan 7 5 ^{o}} - \frac{3 tan 4 5 ^{o} . tan 2 0 ^{o} . tan 4 0 ^{o} . tan 5 0 ^{o} . tan 7 0 ^{o}}{5}

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Text solutionVerified

We have

\frac{2 sin 6 8 ^{o}}{cos 2 2 ^{o}} - \frac{2 cot 1 5 ^{o}}{5 tan 7 5 ^{o}} - \frac{3 tan 4 5 ^{o} . tan 2 0 ^{o} . tan 4 0 ^{o} . tan 5 0 ^{o} . tan 7 0 ^{o}}{5}

= \frac{2 sin ( 9 0 ^{o} - 2 2 ^{o} )}{cos 2 2 ^{o}} - \frac{2 cot 1 5 ^{o}}{5 tan ( 9 0 ^{o} - 1 5 ^{o} )} - \frac{3 tan 4 5 ^{o} . tan 2 0 ^{o} . tan 4 0 ^{o} . tan ( 9 0 ^{o} - 4 0 ^{o} ) . tan ( 9 0 ^{o} - 2 0 ^{o} )}{5}

= \frac{2 cos 2 2 ^{o}}{cos 2 2 ^{o}} - \frac{2 cot 1 5 ^{o}}{5 cot 1 5 ^{o}} - \frac{3 tan 4 5 ^{o} . tan 2 0 ^{o} . tan 4 0 ^{o} . cot 4 0 ^{o} . cot 2 0 ^{o}}{5}

= 2 - \frac{2}{5} - \frac{3 tan 4 5 ^{o} . ( tan 2 0 ^{o} . cot 2 0 ^{o} ) . ( tan 4 0 ^{o} . cot 4 0 ^{o} )}{5}

= 2 - \frac{2}{5} - \frac{3}{5} .1.1.1 = 2 - \frac{2}{5} - \frac{3}{5} = 2 - 1 = 1

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Question Text	Without using tables, evaluate the following: $\frac{2 sin 6 8 ^{o}}{cos 2 2 ^{o}} - \frac{2 cot 1 5 ^{o}}{5 tan 7 5 ^{o}} - \frac{3 tan 4 5 ^{o} . tan 2 0 ^{o} . tan 4 0 ^{o} . tan 5 0 ^{o} . tan 7 0 ^{o}}{5}$
Answer Type	Text solution:1
Upvotes	150

Without using tables, evaluate the following:cos22o2sin68o​−5tan75o2cot15o​−53tan45o.tan20o.tan40o.tan50o.tan70o​

Text solutionVerified

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Without using tables, evaluate the following:
$\frac{2 sin 6 8 ^{o}}{cos 2 2 ^{o}} - \frac{2 cot 1 5 ^{o}}{5 tan 7 5 ^{o}} - \frac{3 tan 4 5 ^{o} . tan 2 0 ^{o} . tan 4 0 ^{o} . tan 5 0 ^{o} . tan 7 0 ^{o}}{5}$