Solve tan-4-1-2cos -11- b -tan-4-1-2cos -1 a - b A- a- BB- s - CC- 2 a -2- bD- 2 b - a

The correct option is D 2b/atan(π/4+1/2cos^ 1a/b )+tan(π/4 1/2cos^ 1a/b ) tanA+tanB=sin(A+B)/cos A.cosB =sin (π/4+π/4+0 )/cos(π/4+π/2cos^ 1a/b ).cos(π/4 π/2co ...

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

Byju's Answer

Standard IX

Mathematics

Use of Trigonometric Table

Solve tanπ/4+...

Question

Solve $t a n (\frac{π}{4} + \frac{1}{2} c o s^{- 1} \frac{a}{b}) + t a n (\frac{π}{4} - \frac{1}{2} c o s^{- 1} \frac{a}{b})$

\frac{b}{a}

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

\frac{a}{b}

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

\frac{2 a}{b}

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

\frac{2 b}{a}

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

Open in App

Solution

The correct option is D $\frac{2 b}{a}$
$t a n (\frac{π}{4} + \frac{1}{2} c o s^{- 1} \frac{a}{b}) + t a n (\frac{π}{4} - \frac{1}{2} c o s^{- 1} \frac{a}{b})$
$t a n A + t a n B = \frac{s i n (A + B)}{c o s A . c o s B}$
$= \frac{s i n (\frac{π}{4} + \frac{π}{4} + 0)}{c o s (\frac{π}{4} + \frac{π}{2} c o s^{- 1} \frac{a}{b}) . c o s (\frac{π}{4} - \frac{π}{2} c o s^{- 1} \frac{a}{b})}$
$= c o s A . c o s B = \frac{1}{2} [c o s (A + B) + c o s (A_{B})]$
$= \frac{s i n \frac{π}{2}}{\frac{1}{2} (\frac{π}{4} + \frac{π}{4} + c o s (\frac{1}{2} c o s^{- 1} \frac{a}{b} + \frac{1}{2} c o s^{- 1} \frac{a}{b}))}$
$= \frac{1}{\frac{1}{2} [0 + c o s c o s^{- 1} \frac{a}{b}]}$
$= \frac{1}{\frac{1}{2} (\frac{a}{b})}$
$= \frac{2 b}{a}$ .

Suggest Corrections

Similar questions

tan (\frac{π}{4} + \frac{1}{2} {cos}^{- 1} \frac{a}{b}) + tan (\frac{π}{4} - \frac{1}{2} {cos}^{- 1} \frac{a}{b}) = \frac{2 b}{a}

t a n (\frac{π}{4} + \frac{1}{2} c o s^{- 1} \frac{a}{b}) + t a n (\frac{π}{4} - \frac{1}{2} c o s^{- 1} \frac{a}{b})

tan (\frac{π}{4} + \frac{1}{2} {cos}^{- 1} \frac{a}{b}) + tan (\frac{π}{4} - \frac{1}{2} {cos}^{- 1} \frac{a}{b})

is equal to

Q. Find

m

tan (\frac{π}{4} + \frac{1}{2} {cos}^{- 1} \frac{a}{b}) + tan (\frac{π}{4} - \frac{1}{2} {cos}^{- 1} \frac{a}{b}) = \frac{m b}{a}

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

t a n (\frac{π}{4} + \frac{1}{2} {c o s}^{- 1} \frac{a}{b}) + t a n (\frac{π}{4} - \frac{1}{2} {c o s}^{- 1} \frac{a}{b}) = \frac{2 b}{a}

Join BYJU'S Learning Program

Solve tan(π4+12cos−1ab)+tan(π4−12cos−1ab)

The correct option is D 2batan(π4+12cos−1ab)+tan(π4−12cos−1ab) tanA+tanB=sin(A+B)cosA.cosB =sin(π4+π4+0)cos(π4+π2cos−1ab).cos(π4−π2cos−1ab) =cosA.cosB=12[cos(A+B)+cos(AB)] =sinπ212(π4+π4+cos(12cos−1ab+12cos−1ab)) =112[0+coscos−1ab] =112(ab) =2ba.

Solve $t a n (\frac{π}{4} + \frac{1}{2} c o s^{- 1} \frac{a}{b}) + t a n (\frac{π}{4} - \frac{1}{2} c o s^{- 1} \frac{a}{b})$