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Question Number 107470 by bemath last updated on 11/Aug/20

$$\circlearrowleft \mathcal{B}e\mathcal{M}ath\circlearrowright$$$$\left(1\right)(1+\tan 3°)(1+\tan 4°)(1+\tan 41°)(1+\tan 42°)=?$$$$\left(2\right)f\left(x\right)=g\left(h\left(x\right)\right);h\left(x\right)=2x^2-3x.$$$$Iff{}^{\prime }(-1)=14theng{}^{\prime }\left(5\right)=?$$

Answered by john santu last updated on 11/Aug/20

$$\divideontimes \mathcal{JS}\divideontimes$$$$\left(2\right)\frac{df\left(x\right)}{dx}=\frac{d(g\left(h\left(x\right)\right)}{dx}$$$$\Rightarrow f{}^{\prime }\left(x\right)=h{}^{\prime }\left(x\right)g{}^{\prime }\left(h\left(x\right)\right)$$$$\Rightarrow f{}^{\prime }(-1)=h^{\prime }(-1)g^{\prime }\left(h(-1)\right)$$$$\bullet \bullet h(-1)=2{(-1)}^2-3(-1)=5$$$$\bullet \bullet g^{\prime }\left(h(-1)\right)=g^{\prime }\left(5\right)$$$$\bullet \bullet h^{\prime }(-1)=4(-1)-3=-7$$$$\Rightarrow f{}^{\prime }(-1)=h^{\prime }(-1)g^{\prime }\left(5\right)$$$$\Rightarrow 14=-7\times g^{\prime }\left(5\right)\iff g{}^{\prime }\left(5\right)=-2$$

Answered by $@y@m last updated on 11/Aug/20

$$(1+\tan 3)(1+\tan 42)=1+\tan 3+\tan 42+\tan 42\tan 3$$$$=1+(\tan 45-\tan 42\tan 3)+\tan 42\tan 3$$$$=2$$$$\therefore thegivenexpession$$$$=2\times 2$$$$=4$$

Commented by bemath last updated on 11/Aug/20

$$thankyousir.$$$$butthequestiononly(1+\tan 3°)(1+\tan 4°)(1+\tan 42°)(1+\tan 41°)$$$$sodoesmeantheresult4sir?$$

Commented by $@y@m last updated on 11/Aug/20

$$Oh.$$$$Yes.$$

Answered by 1549442205PVT last updated on 12/Aug/20

$$\mathrm{We}\mathrm{have}:\mathrm{f}{}^{\prime }\left(\mathrm{x}\right)=\mathrm{g}{}^{\prime }\left(\mathrm{h}\left(\mathrm{x}\right)\right).{\mathrm{h}}^{\prime }\left(\mathrm{x}\right)$$$$\mathrm{f}{}^{\prime }(-1)={\mathrm{g}}^{\prime }\left(\mathrm{h}(-1)\right).{\mathrm{h}}^{\prime }(-1)=14$$$$\Rightarrow {\mathrm{g}}^{\prime }\left(\mathrm{h}(-1)\right)=\frac{14}{-7}=-2\iff {\mathrm{g}}^{\prime }\left(5\right)=-2$$$$\mathrm{b})1+\tan 42=\tan 87(1-\tan 42)=\cot 3(1-\tan 42)$$$$=\frac{1-\tan 42}{\tan 3}\Rightarrow \tan 3(1+\tan 42)=1-\tan 42$$$$\Rightarrow \tan 3=\frac{1-\tan 42}{1+\tan 42}\Rightarrow 1+\tan 3=\frac{2}{1+\tan 42}$$$$1+\tan 41=\tan 86(1-\tan 41)=\cot 4(1-\tan 41)$$$$\Rightarrow 1+\tan 4=\frac{2}{1+\tan 41}$$$$\Rightarrow (1+\tan 3)(1+\tan 4)=\frac{4}{(1+\tan 41)(1+\tan 42)}$$$$\Rightarrow (1+\tan 3)(1+\tan 4)(1+\tan 41)(1+\tan 42)=4$$

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