In ABC- if sin 2 A -sin 2 B but - A - B and 3 tan A -4-0- then value of expression 1-1-tan 2 B -sin B - A -2-Ccos B -1-sin 2 A-cos A -1-sin 2 B isB- 0C-D-

The correct option is C 1/2sin2A=sin2B=sin(π 2B) ⇒2A=π 2B → A+B=π/2⇒ C=π/2 Given that tan A=4/3 Then sin A=4/5, cos A=3/5 ∵ B=π/2 A⇒sin B=3/5, cos B=4/5 Now pu ...

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Byju's Answer

Standard XII

Mathematics

Trigonometric Ratios for Sum of Two Angles

In ABC, if si...

Question

In $△ A B C$ , if $sin 2 A = sin 2 B$ but $∠ A \neq ∠ B$ and $3 tan A - 4 = 0$ , then value of expression $\frac{1}{1 + {tan}^{2} B} + \frac{sin (B - A)}{2} + cot C (cos B \sqrt{1 + {sin}^{2} A} - cos A \sqrt{1 + {sin}^{2} B})$ is

\frac{3}{4}

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\frac{1}{2}

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\frac{1}{4}

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0

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Solution

The correct option is B $\frac{1}{2}$
$sin 2 A = sin 2 B = sin (π - 2 B) \Rightarrow 2 A = π - 2 B \to A + B = \frac{π}{2} \Rightarrow C = \frac{π}{2}$
Given that $tan A = \frac{4}{3}$
Then $sin A = \frac{4}{5}, cos A = \frac{3}{5} ∵ B = \frac{π}{2} - A \Rightarrow sin B = \frac{3}{5}, cos B = \frac{4}{5}$
Now putting all the values in the expression we have the expression as
$\frac{1}{1 + \frac{9}{16}} + \frac{sin B cos A - cos B sin A}{2} + 0 = \frac{16}{25} + \frac{1}{2} (\frac{3}{5} . \frac{3}{5} - \frac{4}{5} . \frac{4}{5}) = \frac{1}{2}$

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Similar questions

Q. In a

△ A B C

the expression

\frac{sin 2 A + sin 2 B + sin 2 C}{sin A + sin B + sin C} = k sin \frac{A}{2} sin \frac{B}{2} sin \frac{C}{2}

, then the value of

k

Q.
lf

A + B + C = 180^{\circ}

. Match the following
List -I List -II

1.

cos A + cos B + cos C

2 + 2 cos A cos B cos C

{sin}^{2} A + {sin}^{2} B + {sin}^{2} C

4 cos \frac{A}{2} cos \frac{B}{2} cos \frac{C}{2}

sin 2 A + sin 2 B + sin 2 C

1 + 4 sin \frac{A}{2} sin \frac{B}{2} sin \frac{C}{2}

sin A + sin B + sin C

4 sin A sin B sin C

Q. If

A + B + C = 0

, (Note

0

\neq π

) then prove that

sin 2 A + sin 2 B + sin 2 C = 2 (sin A + sin B + sin C) (1 + cos A + cos B + cos C)

Q. If

A + B + C = π

then the expression

\frac{sin 2 A + sin 2 B - sin 2 C}{sin 2 A + sin 2 B + sin 2 C}

reduces to

Q. In triangle

A B C

prove that
i.

sin A = sin (B + C)

ii.

sin 2 A = - sin (2 B + 2 C)

iii.

cos A = - cos (A + B)

tan \frac{A + B}{2} = - cot \frac{C}{2}

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In △ABC, if sin2A=sin2B but ∠A≠∠B and 3tanA−4=0, then value of expression 11+tan2B+sin(B−A)2+cotC(cosB√1+sin2A−cosA√1+sin2B) is

The correct option is B 12sin2A=sin2B=sin(π−2B)⇒2A=π−2B→A+B=π2⇒C=π2 Given that tanA=43 Then sinA=45,cosA=35∵B=π2−A⇒sinB=35,cosB=45 Now putting all the values in the expression we have the expression as 11+916+sinBcosA−cosBsinA2+0=1625+12(35.35−45.45)=12

In $△ A B C$ , if $sin 2 A = sin 2 B$ but $∠ A \neq ∠ B$ and $3 tan A - 4 = 0$ , then value of expression $\frac{1}{1 + {tan}^{2} B} + \frac{sin (B - A)}{2} + cot C (cos B \sqrt{1 + {sin}^{2} A} - cos A \sqrt{1 + {sin}^{2} B})$ is