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Question

$I f tan invisible function application x equals n tan invisible function application y comma blank n element of R to the power of plus end exponent comma$ then the maximum value of $s e c to the power of 2 end exponent open parentheses x minus y close parentheses$ is equal to

$\frac{{(n + 1)}^{2}}{2 n}$
$\frac{{(n + 1)}^{2}}{n}$
$\frac{{(n + 1)}^{2}}{2}$
$\frac{{(n + 1)}^{2}}{4 n}$

The correct answer is: $fraction numerator open parentheses n plus 1 close parentheses to the power of 2 end exponent over denominator 4 n end fraction$

$tan invisible function application x equals n tan invisible function application y comma cos invisible function application open parentheses x minus y close parentheses equals cos invisible function application x cos invisible function application y plus sin invisible function application x sin invisible function application y$
$rightwards double arrow cos invisible function application open parentheses x minus y close parentheses equals cos invisible function application x cos invisible function application y open parentheses 1 plus tan invisible function application x. tan invisible function application y close parentheses equals cos invisible function application x cos invisible function application y open parentheses 1 plus n tan to the power of 2 end exponent invisible function application y close parentheses$
$rightwards double arrow sec to the power of 2 end exponent invisible function application open parentheses x minus y close parentheses equals fraction numerator sec to the power of 2 end exponent invisible function application x sec to the power of 2 end exponent invisible function application y over denominator open parentheses 1 plus n tan to the power of 2 end exponent invisible function application y close parentheses to the power of 2 end exponent end fraction$
$equals fraction numerator open parentheses 1 plus tan to the power of 2 end exponent invisible function application x close parentheses open parentheses 1 plus tan to the power of 2 end exponent invisible function application y close parentheses over denominator open parentheses 1 plus n tan to the power of 2 end exponent invisible function application y close parentheses to the power of 2 end exponent end fraction$
$equals fraction numerator open parentheses 1 plus n to the power of 2 end exponent tan to the power of 2 end exponent invisible function application y close parentheses open parentheses 1 plus tan to the power of 2 end exponent invisible function application y close parentheses over denominator open parentheses 1 plus n tan to the power of 2 end exponent invisible function application y close parentheses to the power of 2 end exponent end fraction$
$equals 1 plus fraction numerator open parentheses n minus 1 close parentheses to the power of 2 end exponent tan to the power of 2 end exponent invisible function application y over denominator open parentheses 1 plus n tan to the power of 2 end exponent invisible function application y close parentheses to the power of 2 end exponent end fraction$
$N o w comma blank open parentheses fraction numerator 1 plus n tan to the power of 2 end exponent invisible function application y over denominator 2 end fraction close parentheses greater or equal than n tan to the power of 2 end exponent invisible function application y blank open parentheses because A. M. greater or equal than G. M. close parentheses$
$rightwards double arrow fraction numerator tan to the power of 2 end exponent invisible function application y over denominator open parentheses 1 plus n tan to the power of 2 end exponent invisible function application y close parentheses to the power of 2 end exponent end fraction less or equal than fraction numerator 1 over denominator 4 n end fraction$
$rightwards double arrow sec to the power of 2 end exponent invisible function application open parentheses x minus y close parentheses less or equal than 1 plus fraction numerator open parentheses n minus 1 close parentheses to the power of 2 end exponent over denominator 4 n end fraction equals fraction numerator open parentheses n plus 1 close parentheses to the power of 2 end exponent over denominator 4 n end fraction$

Related Questions to study

If the equation $2 to the power of x end exponent plus 4 to the power of y end exponent equals 2 to the power of y end exponent plus 4 to the power of x end exponent$ is solved for $y$ in terms of $x comma blank$ where $x less than 0 comma$ then the sum of the solutions is

If the equation $2 to the power of x end exponent plus 4 to the power of y end exponent equals 2 to the power of y end exponent plus 4 to the power of x end exponent$ is solved for $y$ in terms of $x comma blank$ where $x less than 0 comma$ then the sum of the solutions is

$I f blank A$ and $blank B$ are acute positive angles satisfying the equations $3 sin to the power of 2 end exponent invisible function application A plus 2 sin to the power of 2 end exponent invisible function application B equals 1$ and $blank 3 sin invisible function application 2 A minus 2 sin invisible function application 2 B equals 0 comma$ then $A plus 2 blank B$ is equal to

$I f blank A$ and $blank B$ are acute positive angles satisfying the equations $3 sin to the power of 2 end exponent invisible function application A plus 2 sin to the power of 2 end exponent invisible function application B equals 1$ and $blank 3 sin invisible function application 2 A minus 2 sin invisible function application 2 B equals 0 comma$ then $A plus 2 blank B$ is equal to

The value of $49 to the power of open parentheses 1 minus log subscript 7 end subscript invisible function application 2 close parentheses end exponent plus 5 to the power of negative log subscript 5 end subscript invisible function application 4 end exponent$ is

The value of $49 to the power of open parentheses 1 minus log subscript 7 end subscript invisible function application 2 close parentheses end exponent plus 5 to the power of negative log subscript 5 end subscript invisible function application 4 end exponent$ is

parallel

If $square root of log subscript 2 end subscript invisible function application x end root minus 0.5 equals log subscript 2 end subscript invisible function application square root of x comma$ then $x$ equals

If $square root of log subscript 2 end subscript invisible function application x end root minus 0.5 equals log subscript 2 end subscript invisible function application square root of x comma$ then $x$ equals

The maximum value of $open parentheses cos invisible function application alpha subscript 1 end subscript close parentheses open parentheses cos invisible function application alpha subscript 2 end subscript close parentheses horizontal ellipsis open parentheses cos invisible function application alpha subscript n end subscript close parentheses comma$ unser the restrictions $0 less or equal than alpha subscript 1 end subscript comma blank alpha subscript 2 end subscript comma horizontal ellipsis alpha subscript n end subscript less or equal than pi divided by 2$ and $open parentheses cot invisible function application alpha subscript 1 end subscript close parentheses open parentheses cot invisible function application alpha subscript 2 end subscript close parentheses horizontal ellipsis open parentheses cot invisible function application alpha subscript n end subscript close parentheses equals 1$ is

The maximum value of $open parentheses cos invisible function application alpha subscript 1 end subscript close parentheses open parentheses cos invisible function application alpha subscript 2 end subscript close parentheses horizontal ellipsis open parentheses cos invisible function application alpha subscript n end subscript close parentheses comma$ unser the restrictions $0 less or equal than alpha subscript 1 end subscript comma blank alpha subscript 2 end subscript comma horizontal ellipsis alpha subscript n end subscript less or equal than pi divided by 2$ and $open parentheses cot invisible function application alpha subscript 1 end subscript close parentheses open parentheses cot invisible function application alpha subscript 2 end subscript close parentheses horizontal ellipsis open parentheses cot invisible function application alpha subscript n end subscript close parentheses equals 1$ is

If both the distinct roots of the equation $open vertical bar sin invisible function application x close vertical bar to the power of 2 end exponent plus open vertical bar sin invisible function application x close vertical bar plus b equals 0 blank$ in $blank open square brackets 0 comma pi close square brackets$ are real, then the values of $b$ are

If both the distinct roots of the equation $open vertical bar sin invisible function application x close vertical bar to the power of 2 end exponent plus open vertical bar sin invisible function application x close vertical bar plus b equals 0 blank$ in $blank open square brackets 0 comma pi close square brackets$ are real, then the values of $b$ are

parallel

In triangle $A B C comma$ internal angle bisector $angle A$ makes an angle $theta$ with side $B C.$ The value of $sin invisible function application theta$ is equal to

In triangle $A B C comma$ internal angle bisector $angle A$ makes an angle $theta$ with side $B C.$ The value of $sin invisible function application theta$ is equal to

If $open parentheses 4 close parentheses to the power of log subscript 9 end subscript invisible function application 3 end exponent plus open parentheses 9 close parentheses to the power of log subscript 2 end subscript invisible function application 4 end exponent equals open parentheses 10 close parentheses to the power of log subscript x end subscript invisible function application 83 end exponent comma$ then $x$ is equal to

If $open parentheses 4 close parentheses to the power of log subscript 9 end subscript invisible function application 3 end exponent plus open parentheses 9 close parentheses to the power of log subscript 2 end subscript invisible function application 4 end exponent equals open parentheses 10 close parentheses to the power of log subscript x end subscript invisible function application 83 end exponent comma$ then $x$ is equal to

The value of $log invisible function application a b minus log invisible function application open vertical bar b close vertical bar equals$

The value of $log invisible function application a b minus log invisible function application open vertical bar b close vertical bar equals$

parallel

If $alpha comma blank beta comma blank gamma comma blank delta$ are the smallest positive angles in ascending order of magnitude which have their sines equal to the positive quantity $k comma$ then the value of $4 sin invisible function application fraction numerator alpha over denominator 2 end fraction plus 3 sin invisible function application fraction numerator beta over denominator 2 end fraction plus 2 sin invisible function application fraction numerator gamma over denominator 2 end fraction plus sin invisible function application fraction numerator delta over denominator 2 end fraction$ is equal to

If $alpha comma blank beta comma blank gamma comma blank delta$ are the smallest positive angles in ascending order of magnitude which have their sines equal to the positive quantity $k comma$ then the value of $4 sin invisible function application fraction numerator alpha over denominator 2 end fraction plus 3 sin invisible function application fraction numerator beta over denominator 2 end fraction plus 2 sin invisible function application fraction numerator gamma over denominator 2 end fraction plus sin invisible function application fraction numerator delta over denominator 2 end fraction$ is equal to

The value of $x$ satisfying $square root of 3 to the power of negative 4 plus 2 log subscript square root of 5 end subscript invisible function application x end exponent equals 1 divided by 9$ is

The value of $x$ satisfying $square root of 3 to the power of negative 4 plus 2 log subscript square root of 5 end subscript invisible function application x end exponent equals 1 divided by 9$ is

In $blank increment A B C$ , if $b to the power of 2 end exponent plus c to the power of 2 end exponent equals 2 a to the power of 2 end exponent comma$ then value of $fraction numerator cot invisible function application A over denominator cot invisible function application B plus cot invisible function application C end fraction$ is

In $blank increment A B C$ , if $b to the power of 2 end exponent plus c to the power of 2 end exponent equals 2 a to the power of 2 end exponent comma$ then value of $fraction numerator cot invisible function application A over denominator cot invisible function application B plus cot invisible function application C end fraction$ is

parallel

If $log subscript 10 end subscript invisible function application open square brackets fraction numerator 1 over denominator 2 to the power of x end exponent plus blank x minus 1 end fraction close square brackets equals x open square brackets log subscript 10 end subscript invisible function application 5 minus 1 close square brackets comma blank t h e n blank x equals$

If $log subscript 10 end subscript invisible function application open square brackets fraction numerator 1 over denominator 2 to the power of x end exponent plus blank x minus 1 end fraction close square brackets equals x open square brackets log subscript 10 end subscript invisible function application 5 minus 1 close square brackets comma blank t h e n blank x equals$

r triangle $A B C comma blank R equals 5 divided by 2 blank$ and $blank r equals 1.$ Let $I$ be the incentre of the triangle and $D comma blank E$ and $blank F$ be the feet of the perpendiculars from $I blank$ to $blank B C comma blank C A blank$ and $blank A B comma$ respectively. The value of $fraction numerator I D cross times I E cross times I F over denominator I A cross times I B cross times I C end fraction blank$ is equal to

r triangle $A B C comma blank R equals 5 divided by 2 blank$ and $blank r equals 1.$ Let $I$ be the incentre of the triangle and $D comma blank E$ and $blank F$ be the feet of the perpendiculars from $I blank$ to $blank B C comma blank C A blank$ and $blank A B comma$ respectively. The value of $fraction numerator I D cross times I E cross times I F over denominator I A cross times I B cross times I C end fraction blank$ is equal to

any $increment A B C comma$ the value of $a open parentheses b to the power of 2 end exponent plus c to the power of 2 end exponent close parentheses cos invisible function application A plus b left parenthesis c to the power of 2 end exponent plus a to the power of 2 end exponent right parenthesis cos invisible function application B plus c left parenthesis a to the power of 2 end exponent plus b to the power of 2 end exponent right parenthesis cos invisible function application C equals$

any $increment A B C comma$ the value of $a open parentheses b to the power of 2 end exponent plus c to the power of 2 end exponent close parentheses cos invisible function application A plus b left parenthesis c to the power of 2 end exponent plus a to the power of 2 end exponent right parenthesis cos invisible function application B plus c left parenthesis a to the power of 2 end exponent plus b to the power of 2 end exponent right parenthesis cos invisible function application C equals$

parallel

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