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Question

triangle $A B C comma fraction numerator sin invisible function application A plus sin invisible function application B plus sin invisible function application C over denominator sin invisible function application A plus sin invisible function application B minus sin invisible function application C end fraction$ is equal to

$\tan \frac{A}{2} \cot \frac{B}{2}$
$\cot \frac{A}{2} \tan \frac{B}{2}$
$\cot \frac{A}{2} \cot \frac{B}{2}$
$\tan \frac{A}{2} \tan \frac{B}{2}$

The correct answer is: $cot invisible function application fraction numerator A over denominator 2 end fraction cot invisible function application fraction numerator B over denominator 2 end fraction$

$D to the power of r end exponent equals sin invisible function application A plus sin invisible function application B minus sin invisible function application C$
$equals 2 sin invisible function application fraction numerator A plus B over denominator 2 end fraction cos invisible function application fraction numerator A minus B over denominator 2 end fraction minus 2 sin invisible function application fraction numerator C over denominator 2 end fraction cos invisible function application fraction numerator C over denominator 2 end fraction$
$equals 2 cos invisible function application fraction numerator C over denominator 2 end fraction open square brackets cos invisible function application open parentheses fraction numerator A minus B over denominator 2 end fraction close parentheses minus sin invisible function application fraction numerator C over denominator 2 end fraction close square brackets$
$equals 2 cos invisible function application fraction numerator C over denominator 2 end fraction open square brackets cos invisible function application fraction numerator A minus B over denominator 2 end fraction minus cos invisible function application fraction numerator A plus B over denominator 2 end fraction close square brackets$
$equals 2 cos invisible function application fraction numerator C over denominator 2 end fraction open square brackets 2 sin invisible function application fraction numerator A over denominator 2 end fraction sin invisible function application fraction numerator B over denominator 2 end fraction close square brackets$
$equals 4 sin invisible function application fraction numerator A over denominator 2 end fraction sin invisible function application fraction numerator B over denominator 2 end fraction cos invisible function application fraction numerator C over denominator 2 end fraction$
$A l s o sin invisible function application A plus sin invisible function application B plus sin invisible function application C equals 4 cos invisible function application fraction numerator A over denominator 2 end fraction cos invisible function application fraction numerator B over denominator 2 end fraction cos invisible function application fraction numerator C over denominator 2 end fraction$
$rightwards double arrow fraction numerator sin invisible function application A plus sin invisible function application B plus sin invisible function application C over denominator sin invisible function application A plus sin invisible function application B minus sin invisible function application C end fraction equals cot invisible function application fraction numerator A over denominator 2 end fraction cot invisible function application fraction numerator B over denominator 2 end fraction$

Related Questions to study

e total number of solutions of $ln invisible function application open vertical bar sin invisible function application x close vertical bar equals negative x to the power of 2 end exponent plus 2 x blank$ in $open square brackets negative fraction numerator pi over denominator 2 end fraction comma fraction numerator 3 pi over denominator 2 end fraction close square brackets$ is equal to

e total number of solutions of $ln invisible function application open vertical bar sin invisible function application x close vertical bar equals negative x to the power of 2 end exponent plus 2 x blank$ in $open square brackets negative fraction numerator pi over denominator 2 end fraction comma fraction numerator 3 pi over denominator 2 end fraction close square brackets$ is equal to

e general solution of the equation $8 cos invisible function application x cos invisible function application 2 x cos invisible function application 4 x equals sin invisible function application 6 x divided by sin invisible function application x$ is

e general solution of the equation $8 cos invisible function application x cos invisible function application 2 x cos invisible function application 4 x equals sin invisible function application 6 x divided by sin invisible function application x$ is

$alpha$ is a root of $25 cos to the power of 2 end exponent invisible function application theta plus 5 cos invisible function application theta minus 12 equals 0 comma fraction numerator pi over denominator 2 end fraction less than alpha less than pi comma$ then $blank sin invisible function application 2 alpha blank$ is equal to

$alpha$ is a root of $25 cos to the power of 2 end exponent invisible function application theta plus 5 cos invisible function application theta minus 12 equals 0 comma fraction numerator pi over denominator 2 end fraction less than alpha less than pi comma$ then $blank sin invisible function application 2 alpha blank$ is equal to

parallel

$x y to the power of 2 end exponent equals 4 blank a n d log subscript 3 end subscript invisible function application open parentheses log subscript 2 end subscript invisible function application x close parentheses plus log subscript 1 divided by 3 end subscript invisible function application open parentheses log subscript 1 divided by 2 end subscript invisible function application y close parentheses equals 1 comma$ then $x$ equals

$x y to the power of 2 end exponent equals 4 blank a n d log subscript 3 end subscript invisible function application open parentheses log subscript 2 end subscript invisible function application x close parentheses plus log subscript 1 divided by 3 end subscript invisible function application open parentheses log subscript 1 divided by 2 end subscript invisible function application y close parentheses equals 1 comma$ then $x$ equals

$I$ is the incentre of a triangle $A B C comma$ then the ratio $I A colon I B colon I C$ is equal to

$I$ is the incentre of a triangle $A B C comma$ then the ratio $I A colon I B colon I C$ is equal to

$log subscript 3 end subscript invisible function application left curly bracket 5 plus 4 log subscript 3 end subscript invisible function application open parentheses x minus 1 close parentheses right curly bracket blank equals 2 comma$ then $x$ is equal to

$log subscript 3 end subscript invisible function application left curly bracket 5 plus 4 log subscript 3 end subscript invisible function application open parentheses x minus 1 close parentheses right curly bracket blank equals 2 comma$ then $x$ is equal to

parallel

$log subscript y end subscript invisible function application x plus log subscript x end subscript invisible function application y equals 1 comma blank x to the power of 2 end exponent plus y equals 12 comma blank$ then the value of $x y$ is

$log subscript y end subscript invisible function application x plus log subscript x end subscript invisible function application y equals 1 comma blank x to the power of 2 end exponent plus y equals 12 comma blank$ then the value of $x y$ is

e number of solutions of the equation $tan invisible function application x plus sec invisible function application x minus 2 cos invisible function application x$ lying in the interval $left square bracket 0 comma blank 2 pi right square bracket$ is

e number of solutions of the equation $tan invisible function application x plus sec invisible function application x minus 2 cos invisible function application x$ lying in the interval $left square bracket 0 comma blank 2 pi right square bracket$ is

a convex quadrilateral $A B C D comma blank A B equals a comma blank B C equals b comma blank C D equals c blank$ and $D A equals d.$ This quadrilateral is such that a circle can be inscribed in it and a circle can be also circumscribed about it, then $tan to the power of 2 end exponent invisible function application open parentheses A divided by 2 close parentheses$ is equal to

a convex quadrilateral $A B C D comma blank A B equals a comma blank B C equals b comma blank C D equals c blank$ and $D A equals d.$ This quadrilateral is such that a circle can be inscribed in it and a circle can be also circumscribed about it, then $tan to the power of 2 end exponent invisible function application open parentheses A divided by 2 close parentheses$ is equal to

parallel

triangle $A B C comma$ line joining the circumcentre and orthocenter is parallel to side $A C comma$ then the value of $tan invisible function application A tan invisible function application C$ is equal to

triangle $A B C comma$ line joining the circumcentre and orthocenter is parallel to side $A C comma$ then the value of $tan invisible function application A tan invisible function application C$ is equal to

$increment A B C comma blank a comma blank b comma blank A$ are given and $c subscript 1 end subscript comma c subscript 2 end subscript$ are two values of the third side $c.$ The sum of the areas of the two triangles with sides $a comma blank b comma blank c subscript 1 end subscript$ and $blank a comma blank b comma blank c subscript 2 end subscript$ is

$increment A B C comma blank a comma blank b comma blank A$ are given and $c subscript 1 end subscript comma c subscript 2 end subscript$ are two values of the third side $c.$ The sum of the areas of the two triangles with sides $a comma blank b comma blank c subscript 1 end subscript$ and $blank a comma blank b comma blank c subscript 2 end subscript$ is

$log subscript 2 end subscript invisible function application x plus log subscript 2 end subscript invisible function application y greater or equal than 6 comma$ then the least value of $x plus y$ is

$log subscript 2 end subscript invisible function application x plus log subscript 2 end subscript invisible function application y greater or equal than 6 comma$ then the least value of $x plus y$ is

parallel

t $a greater than 1$ be a real number. Then the number of roots equation $a to the power of 2 log subscript 2 end subscript invisible function application x end exponent equals 5 plus 4 x to the power of log subscript 2 end subscript invisible function application a end exponent$ has

t $a greater than 1$ be a real number. Then the number of roots equation $a to the power of 2 log subscript 2 end subscript invisible function application x end exponent equals 5 plus 4 x to the power of log subscript 2 end subscript invisible function application a end exponent$ has

of ordered pairs which satisfy the equation $x to the power of 2 end exponent plus 2 x sin invisible function application open parentheses x y close parentheses plus 1 equals 0$ are $blank$ (where $y element of open square brackets 0 comma blank 2 pi close square brackets right parenthesis$

of ordered pairs which satisfy the equation $x to the power of 2 end exponent plus 2 x sin invisible function application open parentheses x y close parentheses plus 1 equals 0$ are $blank$ (where $y element of open square brackets 0 comma blank 2 pi close square brackets right parenthesis$

$increment A B C comma$ if $sin to the power of 2 end exponent invisible function application fraction numerator A over denominator 2 end fraction comma sin to the power of 2 end exponent invisible function application fraction numerator B over denominator 2 end fraction$ and $sin to the power of 2 end exponent invisible function application fraction numerator C over denominator 2 end fraction$ are in H.P., then $blank a comma blank b blank$ and $blank c blank$ will be in

$increment A B C comma$ if $sin to the power of 2 end exponent invisible function application fraction numerator A over denominator 2 end fraction comma sin to the power of 2 end exponent invisible function application fraction numerator B over denominator 2 end fraction$ and $sin to the power of 2 end exponent invisible function application fraction numerator C over denominator 2 end fraction$ are in H.P., then $blank a comma blank b blank$ and $blank c blank$ will be in

parallel

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