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Maths-

General

Easy

Question

$I f blank x comma blank y comma blank z blank a r e blank i n blank A. P comma blank t h e n fraction numerator sin invisible function application x minus sin invisible function application z over denominator cos invisible function application z minus cos invisible function application x end fraction blank i s blank e q u a l blank t o$

$\tan y$
$\cot y$
$\sin y$
$\cos y$

The correct answer is: $cot invisible function application y$

$fraction numerator sin invisible function application x minus sin invisible function application z over denominator cos invisible function application z minus cos invisible function application x end fraction equals fraction numerator 2 cos invisible function application open parentheses fraction numerator x plus z over denominator 2 end fraction close parentheses sin invisible function application open parentheses fraction numerator x minus z over denominator z end fraction close parentheses over denominator 2 sin invisible function application open parentheses fraction numerator x plus z over denominator z end fraction close parentheses sin invisible function application open parentheses fraction numerator x minus z over denominator z end fraction close parentheses end fraction equals cot invisible function application open parentheses fraction numerator x plus z over denominator 2 end fraction close parentheses equals cot invisible function application open parentheses y close parentheses$

Related Questions to study

The general value of $x$ satisfying the equation $2 cot to the power of 2 end exponent invisible function application x plus 2 square root of 3 cot invisible function application x plus 4 blank c o s e c blank x plus 8 equals 0$ is

The general value of $x$ satisfying the equation $2 cot to the power of 2 end exponent invisible function application x plus 2 square root of 3 cot invisible function application x plus 4 blank c o s e c blank x plus 8 equals 0$ is

If $open parentheses 21.4 close parentheses to the power of a end exponent equals open parentheses 0.00214 close parentheses to the power of b end exponent equals 100 comma$ then the value of $fraction numerator 1 over denominator b end fraction minus fraction numerator 1 over denominator b end fraction$ is

If $open parentheses 21.4 close parentheses to the power of a end exponent equals open parentheses 0.00214 close parentheses to the power of b end exponent equals 100 comma$ then the value of $fraction numerator 1 over denominator b end fraction minus fraction numerator 1 over denominator b end fraction$ is

In triangle $A B C comma fraction numerator a over denominator b end fraction equals fraction numerator 2 over denominator 3 end fraction blank$ and $sec to the power of 2 end exponent invisible function application A equals fraction numerator 8 over denominator 5 end fraction.$ Then the number of triangles satisfying these conditions is

In triangle $A B C comma fraction numerator a over denominator b end fraction equals fraction numerator 2 over denominator 3 end fraction blank$ and $sec to the power of 2 end exponent invisible function application A equals fraction numerator 8 over denominator 5 end fraction.$ Then the number of triangles satisfying these conditions is

parallel

The greatest value of $sin to the power of 4 end exponent invisible function application theta plus cos to the power of 4 end exponent invisible function application theta$ is

The greatest value of $sin to the power of 4 end exponent invisible function application theta plus cos to the power of 4 end exponent invisible function application theta$ is

If in a $increment A B C comma cos invisible function application 3 A plus cos invisible function application 3 B plus cos invisible function application 3 C equals 1 comma blank$ then one angle must be exactly equal to

If in a $increment A B C comma cos invisible function application 3 A plus cos invisible function application 3 B plus cos invisible function application 3 C equals 1 comma blank$ then one angle must be exactly equal to

The solution of $4 sin to the power of 2 end exponent invisible function application x plus tan to the power of 2 end exponent invisible function application x plus c o s e c to the power of 2 end exponent x plus cot to the power of 2 end exponent invisible function application x minus 6 equals 0$ is

The solution of $4 sin to the power of 2 end exponent invisible function application x plus tan to the power of 2 end exponent invisible function application x plus c o s e c to the power of 2 end exponent x plus cot to the power of 2 end exponent invisible function application x minus 6 equals 0$ is

parallel

In triangle $A B C comma blank$ if $blank A minus B equals 120 degree blank$ and $blank R equals 8 r$ where $R blank$ and $blank r$ have their usual meaning, then $cos invisible function application C$ equals

In triangle $A B C comma blank$ if $blank A minus B equals 120 degree blank$ and $blank R equals 8 r$ where $R blank$ and $blank r$ have their usual meaning, then $cos invisible function application C$ equals

In the given figure, $A B$ is the diameter of the circle, centered at $´ O to the power of ´ end exponent.$ If $angle C O A equals 60 degree comma blank A B equals 2 r comma blank A C equals d blank$ and $blank C D equals l comma$ then $l$ is equal to

In the given figure, $A B$ is the diameter of the circle, centered at $´ O to the power of ´ end exponent.$ If $angle C O A equals 60 degree comma blank A B equals 2 r comma blank A C equals d blank$ and $blank C D equals l comma$ then $l$ is equal to
Maths-General

$I f tan invisible function application x equals n tan invisible function application y comma blank n element of R to the power of plus end exponent comma$ 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

$I f tan invisible function application x equals n tan invisible function application y comma blank n element of R to the power of plus end exponent comma$ 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

parallel

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

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