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Question

Member of roots of $open parentheses 1 minus tan invisible function application theta close parentheses open parentheses 1 plus sin invisible function application 2 theta close parentheses equals 1 plus tan invisible function application theta$ for $blank theta element of open square brackets 0 comma blank 2 pi close square brackets$ is

$- 1$ and 1
$0 a n d 1$
$1$ and $2$
None of these

The correct answer is: $0 blank a n d blank 1$

$294 blank cos to the power of 2 n end exponent invisible function application x$ lies between
$0 less or equal than cos to the power of 2 n end exponent invisible function application x less or equal than cos to the power of 2 end exponent invisible function application x blank open square brackets a s sin to the power of 4 end exponent invisible function application x equals sin to the power of 2 end exponent invisible function application x sin to the power of 2 end exponent invisible function application x less or equal than sin to the power of 2 end exponent invisible function application x comma sin to the power of 4 end exponent invisible function application x less or equal than sin to the power of 2 end exponent invisible function application x blank a n d blank s o blank o n close square brackets$
$rightwards double arrow 0 less or equal than sin to the power of 2 n end exponent invisible function application x plus cos to the power of 2 n end exponent invisible function application x less or equal than sin to the power of 2 end exponent invisible function application x plus cos to the power of 2 end exponent invisible function application x equals 1$
$rightwards double arrow 0 less or equal than sin to the power of 2 n end exponent invisible function application x plus cos to the power of 2 n end exponent invisible function application x less or equal than 1$
$rightwards double arrow open parentheses 1 minus tan invisible function application theta close parentheses open parentheses 1 plus tan invisible function application theta close parentheses to the power of 2 end exponent equals open parentheses 1 plus tan invisible function application theta close parentheses open parentheses 1 plus tan to the power of 2 end exponent invisible function application theta close parentheses$
$rightwards double arrow open parentheses 1 plus tan invisible function application theta close parentheses open square brackets open parentheses 1 minus tan to the power of 2 end exponent invisible function application theta close parentheses minus open parentheses 1 plus tan to the power of 2 end exponent invisible function application theta close parentheses close square brackets equals 0$
$rightwards double arrow negative 2 tan to the power of 2 end exponent invisible function application theta equals 0 comma blank open parentheses 1 plus tan invisible function application theta close parentheses equals 0$
$rightwards double arrow tan invisible function application theta equals 0$ , or $tan invisible function application theta equals negative 1$
$rightwards double arrow theta equals n pi$ or $blank n pi minus pi divided by 4 comma blank for all blank n element of Z$ , for $blank theta element of open square brackets 0 comma blank 2 pi close square brackets blank theta equals 0 comma blank pi comma blank 2 pi comma blank 3 pi divided by 4 comma blank 7 pi divided by 4$

Related Questions to study

r $blank n element of Z comma blank$ the general solution of $open parentheses square root of 3 minus 1 close parentheses sin invisible function application theta plus open parentheses square root of 3 plus 1 close parentheses cos invisible function application theta equals 2 blank$ is $open parentheses n element of Z close parentheses$

r $blank n element of Z comma blank$ the general solution of $open parentheses square root of 3 minus 1 close parentheses sin invisible function application theta plus open parentheses square root of 3 plus 1 close parentheses cos invisible function application theta equals 2 blank$ is $open parentheses n element of Z close parentheses$

e set of all $x$ in $open parentheses fraction numerator – pi over denominator 2 end fraction comma fraction numerator pi over denominator 2 end fraction close parentheses blank$ satisfying $blank open vertical bar 4 sin invisible function application x minus 1 close vertical bar less than square root of 5$ is given by

e set of all $x$ in $open parentheses fraction numerator – pi over denominator 2 end fraction comma fraction numerator pi over denominator 2 end fraction close parentheses blank$ satisfying $blank open vertical bar 4 sin invisible function application x minus 1 close vertical bar less than square root of 5$ is given by

$x comma y element of open square brackets 0 comma blank 2 pi close square brackets$ and $sin invisible function application x plus sin invisible function application y equals 2$ , then the value of $blank x plus y$ is

$x comma y element of open square brackets 0 comma blank 2 pi close square brackets$ and $sin invisible function application x plus sin invisible function application y equals 2$ , then the value of $blank x plus y$ is

parallel

e value of $open parentheses 1 plus cos invisible function application fraction numerator pi over denominator 8 end fraction close parentheses open parentheses 1 plus cos invisible function application fraction numerator 3 pi over denominator 8 end fraction close parentheses open parentheses 1 plus cos invisible function application fraction numerator 5 pi over denominator 8 end fraction close parentheses open parentheses 1 plus cos invisible function application fraction numerator 7 pi over denominator 8 end fraction close parentheses$ is

e value of $open parentheses 1 plus cos invisible function application fraction numerator pi over denominator 8 end fraction close parentheses open parentheses 1 plus cos invisible function application fraction numerator 3 pi over denominator 8 end fraction close parentheses open parentheses 1 plus cos invisible function application fraction numerator 5 pi over denominator 8 end fraction close parentheses open parentheses 1 plus cos invisible function application fraction numerator 7 pi over denominator 8 end fraction close parentheses$ is

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

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

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

parallel

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

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

parallel

$I$ 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

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

parallel

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

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

parallel

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