The value of s i n [ 2 t a n − 1 ( 1 3 ) ] + c o s [ t a n − 1 ( 2 √ 2 ) ] = Options: A . &nbsp 16 15 B . &nbsp 14 15 C . &nbsp 12 15 D . &nbsp 11 15

Answer: Option B : B s i n [ 2 t a n − 1 ( 1 3 ) ] + c o s [ t a n − 1 ( 2 √ 2 ) ] = s i n [ t a n − 1 2 3 1 − 1 9 ] + c o s [ t a n − 1 ( 2 √ 2 ) ] = s i n [ t a n − 1 3 4 ] + c o s [ t a n − 1 2 √ 2 ] = t a n − 1 2 √ 2 = c o s − 1 1 3 Also t a n − 1 3 4 = s i n − 1 3 5 = 3 5 + 1 3 = 14 15 .

The value of sin[2tan−1(13)]+cos[tan−1(2√2)]=

Discussion Forum : Inverse Trigonometric Functions

Question -

The value of $s i n [2 t a n^{- 1} (\frac{1}{3})] + c o s [t a n^{- 1} (2 \sqrt{2})] =$

Options:

A .

\frac{16}{15}

B .

\frac{14}{15}

C .

\frac{12}{15}

D .

\frac{11}{15}

Answer: Option B
:
B

s i n [2 t a n^{- 1} (\frac{1}{3})] + c o s [t a n^{- 1} (2 \sqrt{2})]

= s i n [t a n^{- 1} \frac{\frac{2}{3}}{1 - \frac{1}{9}}] + c o s [t a n^{- 1} (2 \sqrt{2})]

= s i n [t a n^{- 1} \frac{3}{4}] + c o s [t a n^{- 1} 2 \sqrt{2}]

= t a n^{- 1} 2 \sqrt{2} = c o s^{- 1} \frac{1}{3}

Also

t a n^{- 1} \frac{3}{4} = s i n^{- 1} \frac{3}{5}

= \frac{3}{5} + \frac{1}{3} = \frac{14}{15}

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Discussion Forum : Inverse Trigonometric Functions

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