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Question Number 81963 by jagoll last updated on 17/Feb/20

$$if\tan \left(x\right)+\sec \left(x\right)=\frac{7}{8}$$$$find\cot \left(x\right)+\mathrm{cosec}\left(x\right)=$$

Commented by john santu last updated on 17/Feb/20

$$let\cot \left(x\right)+\mathrm{cosec}\left(x\right)=t$$$$\left(i\right)\{\tan \left(x\right)+\sec \left(x\right)\}\times \{\cot \left(x\right)+\mathrm{cosec}\left(x\right)\}=\frac{7t}{8}$$$$1+\frac{1}{\cos \left(x\right)}+\frac{1}{\sin \left(x\right)}+\frac{1}{\sin \left(x\right)\cos \left(x\right)}=\frac{7t}{8}$$$$\left(ii\right)-\frac{\sin \left(x\right)}{\cos \left(x\right)}-\frac{1}{\cos \left(x\right)}=-\frac{7}{8}$$$$\left(iii\right)-\frac{\cos \left(x\right)}{\sin \left(x\right)}-\frac{1}{\sin \left(x\right)}=-t$$$$\left(i\right)+\left(ii\right)+\left(iii\right)$$$$\Rightarrow 1=-\frac{t}{8}-\frac{7}{8}$$$$\Rightarrow -8=t+7\Rightarrow t=-15$$$$\therefore \cot \left(x\right)+\mathrm{cosec}\left(x\right)=-15$$

Commented by jagoll last updated on 17/Feb/20

$$thankyoumister$$

Answered by Kunal12588 last updated on 17/Feb/20

$$lettanx=t$$$$secx=\sqrt{1+t^2}$$$$cotx=\frac{1}{t}$$$$cscx=\frac{\sqrt{1+t^2}}{t}$$$$t+\sqrt{1+t^2}=\frac{7}{8}$$$$\Rightarrow 1+t^2=\frac{49}{64}+t^2-\frac{7}{4}t$$$$\Rightarrow \frac{7}{4}t=\frac{49}{64}-1=\frac{-15}{64}$$$$\Rightarrow t=\frac{-15}{112}$$$$\sqrt{1+t^2}=\frac{7}{8}-t=\frac{7}{8}+\frac{15}{112}=\frac{98+15}{112}=\frac{113}{112}$$$$\frac{1}{t}+\frac{\sqrt{1+t^2}}{t}=\frac{1+\sqrt{1+t^2}}{t}=\frac{1+\frac{113}{112}}{\frac{-15}{112}}=-\frac{225}{15}=-15$$$$cotx+cscx=-15$$$$Ithinkthereshouldbeonemorevaluewhen$$$$xisinIVquadrant$$

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