Let x be in radians with 0 < x < 2 π . If sin ( 2 sin x ) = cos ( 2 cos x ) ; then tan x + cot x can be written as π c − b a where a , b , c ∈ N . Then the value of ( 25 a + b + c ) is

sin ( 2 sin x ) = sin ( 2 π − 2 cos x ) sin x + cos x = 4 π s.o.b.s 1 + sin 2 x = 16 π 2 sin 2 x = 16 π 2 − 16 ∴ tan x + cot x = s i n 2 x 2 = π 2 − 16 2 × 16 = π 2 − 16 32 ∴ a = 32 , b = 16 , c = 2 25 a + b + c = 2

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Mathematics - Trigonometric Equation Question with Solution | TestHub

MathematicsTrigonometric EquationTrigonometric EquationMedium2 minQB

MathematicsMediuminteger

Let x be in radians with $0 < x < \frac{π}{2}$ . If $sin (2 sin x) = cos (2 cos x)$ ; then $tan x + cot x$ can be written as $\frac{a}{π ^{c} - b}$ where $a, b, c \in N$ . Then the value of $(\frac{a + b + c}{25})$ is

Answer:

Solution:

$sin (2 sin x) = sin (\frac{π}{2} - 2 cos x)$

$sin x + cos x = \frac{π}{4}$

s.o.b.s

$1 + sin 2 x = \frac{π ^{2}}{16}$

$sin 2 x = \frac{π ^{2} - 16}{16}$

$∴ tan x + cot x = \frac{2}{s i n 2 x} = \frac{2 \times 16}{π ^{2} - 16} = \frac{32}{π ^{2} - 16}$

$∴ a = 32, b = 16, c = 2$

$\frac{a + b + c}{25} = 2$

Stream:JEESubject:MathematicsTopic:Trigonometric EquationSubtopic:Trigonometric Equation

⏱ 2mℹ️ Source: QB

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