The value of sin - 2tan-1 1-3 -cos-tan-12-2-

Sin [ 2tan^ 1 ( 1/3 ) ]+cos[tan^ 1(2√(2))] = sin [ tan^ 12/3/1 1/9 ]+cos[tan^ 1(2√(2))] =sin [ tan^ 13/4 ]+cos[tan^ 12√(2)] =tan^ 12√(2) = cos^ 11/3 Also tan ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Properties Derived from Trigonometric Identities

The value of ...

Question

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

\frac{16}{15}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{14}{15}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{12}{15}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{11}{15}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $\frac{14}{15}$
$s i n [2 t a n^{- 1} (\frac{1}{3})] + c o s [t a n^{- 1} (2 \sqrt{2})]$
$= s i n [t a n^{- 1} \frac{\frac{2}{3}}{1 - \frac{1}{9}}] + c o s [t a n^{- 1} (2 \sqrt{2})]$
$= s i n [t a n^{- 1} \frac{3}{4}] + c o s [t a n^{- 1} 2 \sqrt{2}]$
$= t a n^{- 1} 2 \sqrt{2} = c o s^{- 1} \frac{1}{3}$
Also $t a n^{- 1} \frac{3}{4} = s i n^{- 1} \frac{3}{5}$
$= \frac{3}{5} + \frac{1}{3} = \frac{14}{15}$ .

Suggest Corrections

Similar questions

Q. The value of

2 {tan}^{- 1} \frac{1}{3} + {tan}^{- 1} \frac{1}{2} =

Evaluate:

(a)

{sin}^{- 1} \frac{4}{5} + 2 {tan}^{- 1} \frac{1}{3} = \frac{π}{2}

(b)

{tan}^{- 1} \frac{1}{7} + 2 {tan}^{- 1} \frac{1}{3} = \frac{π}{4}

(c)

{tan}^{- 1} \frac{1}{5} + {tan}^{- 1} \frac{1}{7} + {tan}^{- 1} \frac{1}{3} + {tan}^{- 1} \frac{1}{8} = \frac{π}{4}

Q. Prove the following

(1) {sin}^{- 1} (\frac{2 x}{1 + x^{2}}) = 2 {tan}^{- 1} x, | x | \leq 1

(2) {cos}^{- 1} (\frac{1 - x^{2}}{1 + x^{2}}) = 2 {tan}^{- 1} x, x \geq 0

(3) {tan}^{- 1} (\frac{2 x}{1 - x^{2}}) = 2 {tan}^{- 1} x, - 1 < x < 1

Q. Inverse circular functions,Principal values of

{s i n}^{- 1} x, {c o s}^{- 1} x, {t a n}^{- 1} x

{t a n}^{- 1} x + {t a n}^{- 1} y = {t a n}^{- 1} \frac{x + y}{1 - x y}

x y < 1

π + {t a n}^{- 1} \frac{x + y}{1 - x y}

x y > 1

.
Prove that
(a)

2 {t a n}^{- 1} \frac{1}{5} + {s e c}^{- 1} \frac{5 \sqrt{2}}{7} + 2 {t a n}^{- 1} \frac{1}{8} = \frac{π}{4}

(b)

{c o s}^{- 1} \frac{12}{13} + 2 {c o s}^{- 1} \sqrt{(\frac{64}{65})} + {c o s}^{- 1} \sqrt{(\frac{49}{50})} = {c o s}^{- 1} \frac{1}{\sqrt{2}}

Q. Inverse circular functions,Principal values of

{s i n}^{- 1} x, {c o s}^{- 1} x, {t a n}^{- 1} x

{t a n}^{- 1} x + {t a n}^{- 1} y = {t a n}^{- 1} \frac{x + y}{1 - x y}

x y < 1

π + {t a n}^{- 1} \frac{x + y}{1 - x y}

x y > 1

.
(a)

{t a n}^{- 1} \frac{1}{21} + {t a n}^{- 1} \frac{1}{13} + {t a n}^{- 1} (- \frac{1}{8}) = 0

(b)

{t a n}^{- 1} \frac{2}{3} = \frac{1}{2} {t a n}^{- 1} \frac{12}{5}

(c)

{t a n}^{- 1} \frac{1}{4} + {t a n}^{- 1} \frac{2}{9} = \frac{1}{2} {c o s}^{- 1} \frac{3}{5}

Join BYJU'S Learning Program

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

The correct option is B 1415sin[2tan−1(13)]+cos[tan−1(2√2)] =sin[tan−1231−19]+cos[tan−1(2√2)] =sin[tan−134]+cos[tan−12√2] =tan−12√2=cos−113 Also tan−134=sin−135 =35+13=1415.

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