$fraction numerator d over denominator d x end fraction open curly brackets Tan to the power of negative 1 end exponent invisible function application open parentheses fraction numerator square root of 1 plus x to the power of 2 end exponent end root minus 1 over denominator x end fraction close parentheses close curly brackets equals$

If $theta$ is the angle between two adjacent sides of a parallelogram and p, q are the distances between the parallel sides, then the area of the parallelogram

$fraction numerator d over denominator d x end fraction open parentheses fraction numerator a x plus b over denominator c x plus d end fraction close parentheses equals$

For such questions, we should know how to use u by v method.

$fraction numerator d over denominator d x end fraction open parentheses fraction numerator a x plus b over denominator c x plus d end fraction close parentheses equals$

For such questions, we should know how to use u by v method.

$fraction numerator d over denominator d x end fraction open parentheses e to the power of x to the power of 2 end exponent end exponent a to the power of x end exponent close parentheses equals$

For such questions, we should know u.v method.

$fraction numerator d over denominator d x end fraction open parentheses e to the power of x to the power of 2 end exponent end exponent a to the power of x end exponent close parentheses equals$

For such questions, we should know u.v method.

$fraction numerator d over denominator d x end fraction open parentheses Tan to the power of negative 1 end exponent invisible function application square root of fraction numerator 1 minus c o s invisible function application x over denominator 1 plus c o s invisible function application x end fraction end root close parentheses equals$

We should know different formulas to solve such questions.

$fraction numerator d over denominator d x end fraction open parentheses Tan to the power of negative 1 end exponent invisible function application square root of fraction numerator 1 minus c o s invisible function application x over denominator 1 plus c o s invisible function application x end fraction end root close parentheses equals$

We should know different formulas to solve such questions.

$fraction numerator d over denominator d x end fraction open parentheses Tan to the power of negative 1 end exponent invisible function application open parentheses fraction numerator a minus x over denominator 1 plus infinity x end fraction close parentheses close parentheses equals$

For such questions, we should know different formulas.

$fraction numerator d over denominator d x end fraction open parentheses Tan to the power of negative 1 end exponent invisible function application open parentheses fraction numerator a minus x over denominator 1 plus infinity x end fraction close parentheses close parentheses equals$

For such questions, we should know different formulas.

$fraction numerator d over denominator d x end fraction open parentheses cosec to the power of negative 1 end exponent invisible function application open parentheses e to the power of 2 x plus 1 end exponent close parentheses close parentheses equals$

For such questions, we should know different formulas.

$fraction numerator d over denominator d x end fraction open parentheses cosec to the power of negative 1 end exponent invisible function application open parentheses e to the power of 2 x plus 1 end exponent close parentheses close parentheses equals$

For such questions, we should know different formulas.

$fraction numerator d over denominator d x end fraction open parentheses fraction numerator sin invisible function application x over denominator 1 plus c o s invisible function application x end fraction close parentheses equals$

The alternate method to solve this will be using u by method. It is method used in differentiation when we have a condition of numerator and denominator.

$fraction numerator d over denominator d x end fraction open parentheses fraction numerator sin invisible function application x over denominator 1 plus c o s invisible function application x end fraction close parentheses equals$

The alternate method to solve this will be using u by method. It is method used in differentiation when we have a condition of numerator and denominator.

Locus of the foot of perpendicular fr om (-2,3) to a variable line with $x$ intercept 2 is

Vertices of a triangle are (4, 3) , $left parenthesis 5 blank s i n blank capital theta comma blank 5 blank c o s blank capital theta right parenthesis comma$ $left parenthesis 5 blank c o s blank capital theta comma blank minus 5 blank s i n blank capital theta right parenthesis$ Then the locus of the orthocenter is

A line through A(-5, -4) meets the line x+3y+2=0, 2x+y+4=0 and x-y-5=0 at $B comma$ $C$ and $D$ respectively. If $left parenthesis fraction numerator 15 over denominator A B end fraction right parenthesis to the power of 2 end exponent plus left parenthesis fraction numerator 10 over denominator A C end fraction right parenthesis to the power of 2 end exponent equals left parenthesis fraction numerator 6 over denominator A D end fraction right parenthesis to the power of 2 end exponent$ then the equation of the line is

If $x subscript 1 end subscript comma x subscript 2 end subscript comma x subscript 3 end subscript$ as well as $y subscript 1 end subscript comma y subscript 2 end subscript comma y subscript 3 end subscript$ are in G.P with the same common ratio, then the points $left parenthesis x subscript 1 end subscript comma y subscript 1 end subscript right parenthesis comma$ $left parenthesis x subscript 2 end subscript comma y subscript 2 end subscript right parenthesis comma left parenthesis x subscript 3 end subscript comma y subscript 3 end subscript right parenthesis$ .

The range of $alpha$ for which the points $left parenthesis alpha comma 2 plus alpha right parenthesis$ and $left parenthesis fraction numerator 3 alpha over denominator 2 end fraction comma alpha to the power of 2 end exponent right parenthesis$ lie on opposite side of the line 2x+3y=6 is

The internal bisectors of the a $capital delta A B C$ , having the sides BC=3, CA=5 and AB =4 meet the sides BC, CA and AB in D, E and F respectively, the area of $capital delta D E F$ is

A straight line $L$ with negative slope passes through the point (8, 2) and cuts the positive coordinate axes at points P and Q As $L$ varies, the absolute minimum value of OP+OQ= where $O$ is the origin