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Question-14591




Question Number 14591 by Tinkutara last updated on 02/Jun/17
Answered by prakash jain last updated on 02/Jun/17
For x∈R  cosec^2 x≥1  tan^(100) x≥0  for the equality to hold  cosec x=±1  tan x=0  No real solutions
$$\mathrm{For}\:{x}\in\mathbb{R} \\ $$$$\mathrm{cosec}^{\mathrm{2}} {x}\geqslant\mathrm{1} \\ $$$$\mathrm{tan}^{\mathrm{100}} {x}\geqslant\mathrm{0} \\ $$$$\mathrm{for}\:\mathrm{the}\:\mathrm{equality}\:\mathrm{to}\:\mathrm{hold} \\ $$$$\mathrm{cosec}\:{x}=\pm\mathrm{1} \\ $$$$\mathrm{tan}\:{x}=\mathrm{0} \\ $$$$\mathrm{No}\:\mathrm{real}\:\mathrm{solutions} \\ $$
Commented by mrW1 last updated on 02/Jun/17
the question is probably   sec^(100)  x + tan^(100)  x = 1
$${the}\:{question}\:{is}\:{probably}\: \\ $$$$\mathrm{sec}^{\mathrm{100}} \:{x}\:+\:\mathrm{tan}^{\mathrm{100}} \:{x}\:=\:\mathrm{1} \\ $$
Commented by Tinkutara last updated on 03/Jun/17
Thanks Sir! The answer is also no  solutions exist.
$$\mathrm{Thanks}\:\mathrm{Sir}!\:\mathrm{The}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{also}\:\mathrm{no} \\ $$$$\mathrm{solutions}\:\mathrm{exist}. \\ $$
Commented by mrW1 last updated on 03/Jun/17
if the question is   sec^(100)  x + tan^(100)  x = 1  the solution is  sec x=±1 and tan x=0  ⇒x=nπ
$${if}\:{the}\:{question}\:{is}\: \\ $$$$\mathrm{sec}^{\mathrm{100}} \:{x}\:+\:\mathrm{tan}^{\mathrm{100}} \:{x}\:=\:\mathrm{1} \\ $$$${the}\:{solution}\:{is} \\ $$$$\mathrm{sec}\:{x}=\pm\mathrm{1}\:{and}\:\mathrm{tan}\:{x}=\mathrm{0} \\ $$$$\Rightarrow{x}={n}\pi \\ $$
Commented by Tinkutara last updated on 03/Jun/17
Thanks!
$$\mathrm{Thanks}! \\ $$

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