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Area between Two Curves

Let 2 sin x...

Question

Let $2 sin x + 3 cos y = 3 and 3 sin y + 2 cos x = 4$ then

A

x + y = (4 n + 1) π / 2, n \in I

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B

x + y = (2 n + 1) π / 2, n \in I

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C

x

and

y

can be the two non right angles of a

3 - 4 - 5

triangle with

x > y .

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D

x

and

y

can be the two non right angles of a

3 - 4 - 5

triangle with

y > x

.

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Solution

The correct option is A $x + y = (4 n + 1) π / 2, n \in I$
$2 sin x + 3 cos y = 3$
$3 sin y + 2 cos x = 4$
on squaring and adding, we get
$\Rightarrow 4 + 9 + 12 (sin (x + y)) = 9 + 16$
so, $sin (x + y) = 1$
thus, $x + y = (4 n + 1) \frac{π}{2} \forall n \in I$
Hence, the answer is $x + y = (4 n + 1) \frac{π}{2}, n \in I .$

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Similar questions

\to b

\to c

are non-zero and non-collinear vectors,

\to a \times (\to b \times \to c) + (\to a \cdot \to b) \to b

= (4 - 2 x - sin y) \to b + (x^{2} - 1) \to c

(\to c \cdot \to c) \to a = \to c

Q. Find the angles between each of the following pairs of straight lines:
(i) 3x + y + 12 = 0 and x + 2y − 1 = 0
(ii) 3x − y + 5 = 0 and x − 3y + 1 = 0
(iii) 3x + 4y − 7 = 0 and 4x − 3y + 5 = 0
(iv) x − 4y = 3 and 6x − y = 11
(v) (m² − mn) y = (mn + n²) x + n³ and (mn + m²) y = (mn − n²) x + m³.

\to b

\to c

are non-colinear, if

\to a \times (\to b \times \to c) + (\to a \cdot \to b) \to b = (4 - 2 x - sin y) \to b + (x^{2} - 1) \to c

(\to c \cdot \to c) \to a = \to c

A (x_{1}, y_{1}), B (x_{2}, y_{2}), C (x_{3}, y_{3})

are the vertices of an equilateral triangle and

(x_{1} - 4)^{2} + (y_{1} - 5)^{2} = (x_{2} - 4)^{2} + (y_{2} - 5)^{2} = (x_{3} - 4)^{2} + (y_{3} - 5)^{2},

then the value of

(y_{1} + y_{2} + y_{3}) - (x_{1} + x_{2} + x_{3})

Find the angle between each of the following pairs of straight lines:

(i) $3 x + y + 12 = 0$ and $x + 2 y - 1 = 0$

(ii) $3 x - y + 5 = 0$ and $x - 3 y + 1 = 0$

(iii) $3 x + 4 y - 7 = 0$ and $4 x - 3 y + 5 = 0$

(iv) $x - 4 y = 3$ and $6 x - y = 11$

(v) $(m^{2} - m n) y = (m n + n^{2}) x + n^{2}$ and $(m n + m^{2}) y = (m n + n^{2}) x + m^{3}$ .

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