Have a query? Ask a Tutor!!

Access to best educators and exclusive paper discussions

Offsite Campus Turito Championship

Can’t find what you’re looking for?

Our tutors are from top schools around the world, and they are waiting for you 24/7, anytime, anywhere.

Homework- One to One Solutions
Understand concepts to solve better
Get your doubts solved instantly

Maths-

General

Easy

Question

Sum of all the solutions in $open square brackets 0 comma blank 4 pi close square brackets$ of the equation $tan invisible function application x plus cot invisible function application x plus 1 equals cos invisible function application open parentheses x plus fraction numerator pi over denominator 4 end fraction close parentheses$ is

$3 π$
$π / 2$
$7 π / 2$
$4 π$

The correct answer is: $7 pi divided by 2$

$tan invisible function application x plus cot invisible function application x plus 1 equals cos invisible function application open parentheses x plus fraction numerator pi over denominator 4 end fraction close parentheses$
$rightwards double arrow tan invisible function application x plus cot invisible function application x equals cos invisible function application open parentheses x plus fraction numerator pi over denominator 4 end fraction close parentheses minus 1$
Now $tan invisible function application x plus cot invisible function application x less or equal than negative 2$ and $cos invisible function application open parentheses x plus fraction numerator pi over denominator 4 end fraction close parentheses minus 1 greater or equal than negative 2$
It implies that equality holds when both are $negative 2$
$rightwards double arrow cos invisible function application open parentheses x plus fraction numerator pi over denominator 4 end fraction close parentheses equals negative 1$
$rightwards double arrow x plus fraction numerator pi over denominator 4 end fraction equals open parentheses 2 m plus 1 close parentheses pi comma blank m element of Z$
$rightwards double arrow x equals fraction numerator 3 pi over denominator 4 end fraction blank o r fraction numerator 11 pi over denominator 4 end fraction$
Therefore, the sum of the solutions is $fraction numerator 3 pi over denominator 4 end fraction plus fraction numerator 11 pi over denominator 4 end fraction equals fraction numerator 7 pi over denominator 2 end fraction$

Related Questions to study

If $blank open parentheses 1 minus tan invisible function application theta close parentheses open parentheses 1 plus tan invisible function application theta close parentheses sec to the power of 2 end exponent invisible function application theta plus 2 to the power of tan to the power of 2 end exponent invisible function application theta end exponent equals 0 comma$ then the number of values of $theta$ in the interval $left parenthesis negative pi divided by 2 comma blank pi divided by 2 right parenthesis$ are

If $blank open parentheses 1 minus tan invisible function application theta close parentheses open parentheses 1 plus tan invisible function application theta close parentheses sec to the power of 2 end exponent invisible function application theta plus 2 to the power of tan to the power of 2 end exponent invisible function application theta end exponent equals 0 comma$ then the number of values of $theta$ in the interval $left parenthesis negative pi divided by 2 comma blank pi divided by 2 right parenthesis$ are

If $open vertical bar 2 sin invisible function application theta minus c o s e c blank theta close vertical bar greater or equal than 1$ and $theta not equal to fraction numerator n pi over denominator 2 end fraction comma blank n element of I$ , then

If $open vertical bar 2 sin invisible function application theta minus c o s e c blank theta close vertical bar greater or equal than 1$ and $theta not equal to fraction numerator n pi over denominator 2 end fraction comma blank n element of I$ , then

If $S equals open curly brackets x element of N blank colon 2 plus log subscript 2 end subscript invisible function application square root of x plus 1 end root greater than 1 minus log subscript 1 divided by 2 end subscript invisible function application square root of 4 minus x to the power of 2 end exponent end root close curly brackets comma$ then

If $S equals open curly brackets x element of N blank colon 2 plus log subscript 2 end subscript invisible function application square root of x plus 1 end root greater than 1 minus log subscript 1 divided by 2 end subscript invisible function application square root of 4 minus x to the power of 2 end exponent end root close curly brackets comma$ then

parallel

$log subscript 4 end subscript invisible function application 18$ is

$log subscript 4 end subscript invisible function application 18$ is

If $sin invisible function application 6 theta plus sin invisible function application 4 theta plus sin invisible function application 2 theta equals 0 comma$ then $theta blank$ is equal to $blank open parentheses n element of Z close parentheses$

If $sin invisible function application 6 theta plus sin invisible function application 4 theta plus sin invisible function application 2 theta equals 0 comma$ then $theta blank$ is equal to $blank open parentheses n element of Z close parentheses$

The area of the circle and the area of a regular polygon of $n$ sides and of perimeter equal to that of the circle are in the ratio of

The area of the circle and the area of a regular polygon of $n$ sides and of perimeter equal to that of the circle are in the ratio of

parallel

$I f blank x element of open parentheses pi comma fraction numerator 3 pi over denominator 2 end fraction close parentheses comma blank t h e n blank 4 cos to the power of 2 end exponent invisible function application open parentheses fraction numerator pi over denominator 4 end fraction minus fraction numerator x over denominator 2 end fraction close parentheses plus square root of 4 sin to the power of 4 end exponent invisible function application x plus sin to the power of 2 end exponent invisible function application 2 x end root$ is always equal to

$I f blank x element of open parentheses pi comma fraction numerator 3 pi over denominator 2 end fraction close parentheses comma blank t h e n blank 4 cos to the power of 2 end exponent invisible function application open parentheses fraction numerator pi over denominator 4 end fraction minus fraction numerator x over denominator 2 end fraction close parentheses plus square root of 4 sin to the power of 4 end exponent invisible function application x plus sin to the power of 2 end exponent invisible function application 2 x end root$ is always equal to

If $tan invisible function application theta plus sin invisible function application theta equals m blank a n d tan invisible function application theta minus sin invisible function application theta equals n comma$ then

If $tan invisible function application theta plus sin invisible function application theta equals m blank a n d tan invisible function application theta minus sin invisible function application theta equals n comma$ then

$I f blank x subscript 1 end subscript blank a n d blank x subscript 2 end subscript$ are two distinct roots of the equation $a cos invisible function application x plus b sin invisible function application x equals c comma blank t h e n tan invisible function application fraction numerator x subscript 1 end subscript plus x subscript 2 end subscript over denominator 2 end fraction$ is equal to

$I f blank x subscript 1 end subscript blank a n d blank x subscript 2 end subscript$ are two distinct roots of the equation $a cos invisible function application x plus b sin invisible function application x equals c comma blank t h e n tan invisible function application fraction numerator x subscript 1 end subscript plus x subscript 2 end subscript over denominator 2 end fraction$ is equal to

parallel

$I f fraction numerator sin invisible function application open parentheses x plus y close parentheses over denominator sin invisible function application open parentheses x minus y close parentheses end fraction equals fraction numerator a plus b over denominator a minus b end fraction comma blank t h e n fraction numerator tan invisible function application x over denominator tan invisible function application y end fraction blank i s blank e q u a l blank t o$

$I f fraction numerator sin invisible function application open parentheses x plus y close parentheses over denominator sin invisible function application open parentheses x minus y close parentheses end fraction equals fraction numerator a plus b over denominator a minus b end fraction comma blank t h e n fraction numerator tan invisible function application x over denominator tan invisible function application y end fraction blank i s blank e q u a l blank t o$

Which of the following is not the general solution of $2 to the power of cos invisible function application 2 x end exponent plus 1 equals 3.2 to the power of negative sin to the power of 2 end exponent invisible function application x end exponent ?$

Which of the following is not the general solution of $2 to the power of cos invisible function application 2 x end exponent plus 1 equals 3.2 to the power of negative sin to the power of 2 end exponent invisible function application x end exponent ?$

In triangle $A B C comma$ base $B C$ and area of triangle $increment$ are fixed. Locus of the centroid of triangle $A B C$ is a straight line that is

In triangle $A B C comma$ base $B C$ and area of triangle $increment$ are fixed. Locus of the centroid of triangle $A B C$ is a straight line that is

parallel

The radii $r subscript 1 end subscript comma blank r subscript 2 end subscript comma blank r subscript 3 end subscript$ of the escribed circles of the triangle $A B C$ are in H.P. If the area of the triangle is 24 $c m to the power of 2 end exponent$ and its perimeter is 24 cm, then the length of its largest side is

The radii $r subscript 1 end subscript comma blank r subscript 2 end subscript comma blank r subscript 3 end subscript$ of the escribed circles of the triangle $A B C$ are in H.P. If the area of the triangle is 24 $c m to the power of 2 end exponent$ and its perimeter is 24 cm, then the length of its largest side is

In an acute angled triangle $A B C comma blank r plus r subscript 1 end subscript equals r subscript 2 end subscript plus r subscript 3 end subscript blank$ and $blank angle B greater than fraction numerator pi over denominator 3 end fraction comma$ then

In an acute angled triangle $A B C comma blank r plus r subscript 1 end subscript equals r subscript 2 end subscript plus r subscript 3 end subscript blank$ and $blank angle B greater than fraction numerator pi over denominator 3 end fraction comma$ then

If $´ O ´$ is the circumcentre of $increment A B C blank$ and $blank R subscript 1 end subscript comma R subscript 2 end subscript blank$ and $blank R subscript 3 end subscript$ are the radii of the circumcircles of triangles $O B C comma blank O C A blank$ and $blank O A B comma$ respectively, then $fraction numerator a over denominator R subscript 1 end subscript end fraction plus fraction numerator b over denominator R subscript 2 end subscript end fraction plus fraction numerator c over denominator R subscript 3 end subscript end fraction$ has the value equal to

If $´ O ´$ is the circumcentre of $increment A B C blank$ and $blank R subscript 1 end subscript comma R subscript 2 end subscript blank$ and $blank R subscript 3 end subscript$ are the radii of the circumcircles of triangles $O B C comma blank O C A blank$ and $blank O A B comma$ respectively, then $fraction numerator a over denominator R subscript 1 end subscript end fraction plus fraction numerator b over denominator R subscript 2 end subscript end fraction plus fraction numerator c over denominator R subscript 3 end subscript end fraction$ has the value equal to

parallel

card img

With Turito Academy.

card img

With Turito Foundation.

card img

Get an Expert Advice From Turito.

Turito Academy

card img

With Turito Academy.

Test Prep

card img

With Turito Foundation.