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Answer 1

Answer:

To prove:-

$\bf (1+tan15°)(1+tan30°)=2$

Proof:-

We know that

$\boxed{\bf 15°+30°=45°}$

$\sf{:}\implies tan (15°+30°)=tan45°$

$\sf{:}\implies \dfrac {tan15°+tan30°}{1-tan15°.tan30°}=1$

$\sf{:}\implies tan15°+tan30°=1-tan15°.tan30°$

$\sf{:}\implies tan15°+tan30°+tan15°.tan30°=1$

$\sf{:}\implies 1+tan15°+tan30°+tan15°.tan30°=1+1$

$\sf{:}\implies 1+tan15°+tan30°+tan15°.tan30°=2$

$\sf{:}\implies 1 (1+tan15°)+tan30°(1+tan15°)=2$

$\sf{:}\implies (1+tan15°)(1+tan30°)=2$

$\therefore{\huge{ \bf{(Proved)}}}$

Answer 2

${ \bold { \underline{\large\pink{Correct \: Question :) }}}} \:$

★ Verify the rolle's theorem for the function.

$\\ \displaystyle \sf \: f(x) = e ^{x} \bigg( \sin x - \cos x \bigg) \: on \: \bigg[{\dfrac{\pi}{4} \: \dfrac{5\pi}{4}} \bigg]$

TO PROVE :-

verify rolle's theorem for the given function.

$\Large{\underline{\underline{\textsf{\maltese\: {\red{Required Answer :) }}}}}}$

∵ As we know that the exponential, cosine, sine functions are always continuous and differential in every situation. so we can say that,

$\\ : \implies \displaystyle \sf \: f(x) \: continuous \: on \: x \in \bigg[{\dfrac{\pi}{4} \: \dfrac{5\pi}{4}} \bigg] \: \: \: \: \: \: \: \: \: \: \: \bigg \lgroup Equation \ 1\bigg \rgroup$

$: \implies \displaystyle \sf \: f(x) \: differential \: on \: x \in \bigg[{\dfrac{\pi}{4} \: \dfrac{5\pi}{4}} \bigg] \: \: \: \: \: \: \: \: \: \: \: \bigg \lgroup Equation \ 2\bigg \rgroup$

For x = π/4,

$\\ : \implies \displaystyle \sf \: f(x) = e ^{x} \bigg( \sin x - \cos x \bigg)$

$\\ : \implies \displaystyle \sf \: f \bigg( \dfrac{ \pi}{4} \bigg ) = e ^{x} \bigg( \sin ^{ \dfrac{ \pi}{4} } - \cos ^{ \dfrac{ \pi}{4} } \bigg)$

$: \implies \displaystyle \sf \: f \bigg( \dfrac{ \pi}{4} \bigg ) = e ^{ \dfrac{ \pi}{4} } \bigg( \dfrac{ \sqrt{2} }{2} - \dfrac{\sqrt{2} }{2} \bigg)$

$: \implies \displaystyle \sf \: f \bigg( \dfrac{ \pi}{4} \bigg ) = e ^{ \dfrac{ \pi}{4} } \bigg( \dfrac{ \sqrt{2} - \sqrt{2} }{2} \bigg)$

$: \implies \displaystyle \sf \: f \bigg( \dfrac{ \pi}{4} \bigg ) = e ^{ \dfrac{ \pi}{4} } \bigg( \dfrac{ 0 }{2} \bigg)$

$: \implies \displaystyle \sf \: f \bigg( \dfrac{ \pi}{4} \bigg ) = e ^{ \dfrac{ \pi}{4} } \bigg(0 \bigg)$

$: \implies \displaystyle \sf \: f \bigg( \dfrac{ \pi}{4} \bigg ) = 0$

For x = 5π/4,

$\\ : \implies \displaystyle \sf \: f(x) = e ^{x} \bigg( \sin x - \cos x \bigg)$

$\\ : \implies \displaystyle \sf \: f \bigg( \dfrac{ 5\pi}{4} \bigg ) = e ^{x} \bigg( \sin ^{ \dfrac{5 \pi}{4} } - \cos ^{ \dfrac{5 \pi}{4} } \bigg)$

$: \implies \displaystyle \sf \: f \bigg( \dfrac{5 \pi}{4} \bigg ) = e ^{ \dfrac{ 5\pi}{4} } \bigg( \dfrac{5 \sqrt{2} }{2} - \dfrac{5\sqrt{2} }{2} \bigg)$

$: \implies \displaystyle \sf \: f \bigg( \dfrac{ 5\pi}{4} \bigg ) = e ^{ \dfrac{ 5\pi}{4} } \bigg( \dfrac{5 \sqrt{2} - 5\sqrt{2} }{2} \bigg)$

$: \implies \displaystyle \sf \: f \bigg( \dfrac{5 \pi}{4} \bigg ) = e ^{ \dfrac{ 5\pi}{4} } \bigg( \dfrac{ 0 }{2} \bigg)$

$: \implies \displaystyle \sf \: f \bigg( \dfrac{ 5\pi}{4} \bigg ) = e ^{ \dfrac{ 5\pi}{4} } \bigg(0 \bigg)$

$: \implies \displaystyle \sf \: f \bigg( \dfrac{ 5\pi}{4} \bigg ) = 0$

Now we can say that ,

$\\ : \implies \displaystyle \sf \: f \bigg( \dfrac{ \pi}{4} \bigg ) = f \bigg( \dfrac{5 \pi}{4} \bigg ) \: \: \: \: \: \: \: \: \: \: \: \: \: \bigg \lgroup Equation \ 3\bigg \rgroup$

So now we can say that all the the three equations satisfy for the rolle's theorem.

$\\ : \implies \displaystyle \sf \: f(x) = e ^{x} \bigg( \sin x - \cos x \bigg)$

Now differentiate with respect to x in the given function.

$: \implies \displaystyle \sf \: f' (x) = e ^{x} \bigg( \sin x + \cos x \bigg) + e ^{x} \bigg( \sin x - \cos x \bigg)$

$: \implies \displaystyle \sf \: f' (x) = e ^{x} \bigg( \sin x + \cos x + \sin x - \cos x\bigg )$

$: \implies \displaystyle \sf \: f' (x) = e ^{x} \bigg(2 \sin x \bigg)$

Now,

$\\ : \implies \displaystyle \sf \: f' (x) = 0$

$: \implies \displaystyle \sf \:e ^{x} \bigg(2 \sin x \bigg) = 0$

$: \implies \displaystyle \sf \: \bigg(2 \sin x \bigg) = \frac{0}{e ^{x} }$

$: \implies \displaystyle \sf \: \bigg(2 \sin x \bigg) = 0$

$: \implies \displaystyle \sf \: \sin x = 0$

$: \implies \displaystyle \sf \:x = \pi \: \: on \: x \in \bigg[{\dfrac{\pi}{4} \: \dfrac{5\pi}{4}} \bigg]$

$\Large{\underline{\underline{\textsf{\maltese\: {\orange{Hence~verified}}}}}}$