Suppose α , β and θ are angles satisfying 0 < α < θ < β < 2 π , then c o s β − c o s α s i n α − s i n β =

Let f ( x ) = sin x and g ( x ) = cos x , then f and g are continuous and derivable. Also, sin x  = 0 for any x ∈ ( 0 , 2 π ) so by Cauchy's MVT, g ( β ) − g ( α ) f ( β ) − f ( α ) = g ′ ( θ ) f ′ ( θ ) ⇒ c o s β − c o s α s i n β − s i n α = − s i n θ c o s θ

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Mathematics - Application of Derivative Question with Solution | TestHub

MathematicsApplication of DerivativeMonotonicity-LMVTMedium2 minQB

MathematicsMediumsingle choice

Suppose $α, β$ and $θ$ are angles satisfying $0 < α < θ < β < \frac{π}{2}$ , then $\frac{s i n α - s i n β}{c o s β - c o s α} =$

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Let $f (x) = sin x$ and $g (x) = cos x$ , then f and g are continuous and derivable. Also, $sin x \neq = 0$ for any $x \in (0, \frac{π}{2})$ so by Cauchy's MVT, $\frac{f ( β ) - f ( α )}{g ( β ) - g ( α )} = \frac{f ^{'} ( θ )}{g ^{'} ( θ )} \Rightarrow \frac{s i n β - s i n α}{c o s β - c o s α} = \frac{c o s θ}{- s i n θ}$

Stream:JEESubject:MathematicsTopic:Application of DerivativeSubtopic:Monotonicity-LMVT

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