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If ABC is a triangle and $tan space A over 2 comma tan space B over 2 comma tan space C over 2$ are H.P then the minimum value of $cot space B over 2$ is equal to

$- \sqrt{3}$
$\sqrt{3}$
2
-2

The correct answer is: $square root of 3$

$tan space A over 2 comma tan space B over 2 comma tan space C over 2 space a r e space i n space H. P therefore space tan B over 2 equals fraction numerator 2 tan begin display style A over 2 end style tan begin display style C over 2 end style over denominator tan begin display style A over 2 end style plus tan begin display style C over 2 end style end fraction equals fraction numerator 2 begin display style fraction numerator sin begin display style A over 2 end style over denominator cos begin display style A over 2 end style end fraction end style times begin display style fraction numerator sin begin display style C over 2 end style over denominator cos begin display style C over 2 end style end fraction end style over denominator begin display style fraction numerator sin begin display style A over 2 end style over denominator cos begin display style A over 2 end style end fraction end style plus begin display style fraction numerator sin begin display style C over 2 end style over denominator cos begin display style C over 2 end style end fraction end style end fraction equals fraction numerator 2 sin begin display style A over 2 end style cos begin display style C over 2 end style over denominator sin begin display style A over 2 end style cos begin display style C over 2 end style plus sin begin display style C over 2 end style cos begin display style A over 2 end style end fraction equals fraction numerator cos open parentheses begin display style A over 2 end style minus begin display style C over 2 end style close parentheses minus cos open parentheses begin display style A over 2 end style plus begin display style C over 2 end style close parentheses over denominator sin open parentheses begin display style A over 2 end style plus begin display style C over 2 end style close parentheses end fraction equals fraction numerator cos open parentheses A over 2 minus C over 2 close parentheses minus cos open parentheses begin display style pi over 2 end style minus begin display style B over 2 end style close parentheses over denominator sin open parentheses begin display style pi over 2 end style minus begin display style B over 2 end style close parentheses end fraction tan B over 2 equals fraction numerator cos open parentheses A over 2 minus C over 2 close parentheses minus sin open parentheses begin display style B over 2 end style close parentheses over denominator cos open parentheses begin display style B over 2 end style close parentheses end fraction fraction numerator sin B over 2 over denominator cos begin display style B over 2 end style end fraction equals fraction numerator cos open parentheses A over 2 minus C over 2 close parentheses minus sin open parentheses begin display style B over 2 end style close parentheses over denominator cos open parentheses begin display style B over 2 end style close parentheses end fraction sin B over 2 equals cos open parentheses A over 2 minus C over 2 close parentheses minus sin open parentheses B over 2 close parentheses 2 sin B over 2 equals cos open parentheses A over 2 minus C over 2 close parentheses therefore c o t B over 2 equals square root of 3$

A sequence of numbers is called a harmonic progression if the reciprocal of the terms are in AP. In simple terms, a,b,c are in HP if 1/a, 1/b, 1/c are in AP.

If $theta equals pi over 2 to the power of n plus 1 end exponent$ then $cos space theta cos space 2 theta cos space 2 squared theta horizontal ellipsis cos space 2 to the power of n minus 1 end exponent theta$ is equal to

Maths-General

General

Maths-

If A,B,C re in A.P and $B equals pi over 4$ then $tan space A tan space B tan C$ =

Maths-General

General

maths-

If $alpha comma beta comma gamma$ do not differ by a multiple of $straight pi$ and if $fraction numerator cos space left parenthesis alpha plus theta right parenthesis over denominator sin space left parenthesis beta plus gamma right parenthesis end fraction equals fraction numerator cos begin display style space end style left parenthesis beta plus theta right parenthesis over denominator sin begin display style space end style left parenthesis gamma plus alpha right parenthesis end fraction equals fraction numerator cos begin display style space end style left parenthesis gamma plus theta right parenthesis over denominator sin begin display style space end style left parenthesis alpha plus beta right parenthesis end fraction equals k$ Then k equals

maths-General

General

maths-

From the figure AOC=

maths-General

General

maths-

In $capital delta$ ABC, D,E,F are the midpoints of sides BC, CA and AB respectively then ar $open parentheses B D E F close parentheses equals$

maths-General

General

biology

When the solute has been added in the solution, then following observation can be made ?

biologyGeneral

General

chemistry-

B can be obtained from halide by van Arkel method. This involves reaction

chemistry-General

General

Maths-

If $tan space open parentheses x plus pi over 4 close parentheses equals a$ then $sec squared space x$ =

Maths-General

General

maths-

$sin space pi over 7 sin space fraction numerator 2 pi over denominator 7 end fraction sin space fraction numerator 4 pi over denominator 7 end fraction$ =

maths-General

General

chemistry-

Element $M plus N subscript 2 rightwards arrow with ΔH subscript 2 straight O on top NH subscript 3$ Element M belonging to group 13 can be

chemistry-General

General

maths-

In $blank subscript straight triangle ABC$ $A equals pi over 4 text and end text tan text end text B tan space C equals P n$ $t h e n space a l l space p o s s i b l e space v a l u e s space o f space P space i s$

maths-General

General

maths-

In the given figure, ABCD is a Parallelogram then ar ( $capital delta$ AFB) is

maths-General

General

maths-

In the following figure, if $capital delta$ ABC is circumscribed over the circle, then x =

maths-General

General

maths-

ABC is an Isosceles Triangle inscribed in a Circle such that AB = AC = 17 cm, BC = 16 cm then radius of Circle =

maths-General

General

maths-

In the given fig. if C is the centre of Circle PQC = 25° and PRC =15° then QCR = i

maths-General

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If ABC is a triangle and are H.P then the minimum value of is equal to

2-2

The correct answer is:

Related Questions to study

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If A,B,C re in A.P and then =

If do not differ by a multiple of and if Then k equals

If do not differ by a multiple of and if Then k equals

From the figure AOC=

From the figure AOC=

In ABC, D,E,F are the midpoints of sides BC, CA and AB respectively then ar

In ABC, D,E,F are the midpoints of sides BC, CA and AB respectively then ar

When the solute has been added in the solution, then following observation can be made ?

When the solute has been added in the solution, then following observation can be made ?

B can be obtained from halide by van Arkel method. This involves reaction

B can be obtained from halide by van Arkel method. This involves reaction

If then =

If then =

=

=

Element Element M belonging to group 13 can be

Element Element M belonging to group 13 can be

In

In

In the given figure, ABCD is a Parallelogram then ar (AFB) is

In the given figure, ABCD is a Parallelogram then ar (AFB) is

In the following figure, if ABC is circumscribed over the circle, then x =

In the following figure, if ABC is circumscribed over the circle, then x =

ABC is an Isosceles Triangle inscribed in a Circle such that AB = AC = 17 cm, BC = 16 cm then radius of Circle =

ABC is an Isosceles Triangle inscribed in a Circle such that AB = AC = 17 cm, BC = 16 cm then radius of Circle =

In the given fig. if C is the centre of Circle PQC = 25° and PRC =15° then QCR = i

In the given fig. if C is the centre of Circle PQC = 25° and PRC =15° then QCR = i

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If ABC is a triangle and $tan space A over 2 comma tan space B over 2 comma tan space C over 2$ are H.P then the minimum value of $cot space B over 2$ is equal to

$- \sqrt{3}$
$\sqrt{3}$
2
-2

The correct answer is: $square root of 3$

If $theta equals pi over 2 to the power of n plus 1 end exponent$ then $cos space theta cos space 2 theta cos space 2 squared theta horizontal ellipsis cos space 2 to the power of n minus 1 end exponent theta$ is equal to

If $theta equals pi over 2 to the power of n plus 1 end exponent$ then $cos space theta cos space 2 theta cos space 2 squared theta horizontal ellipsis cos space 2 to the power of n minus 1 end exponent theta$ is equal to

If A,B,C re in A.P and $B equals pi over 4$ then $tan space A tan space B tan C$ =

If A,B,C re in A.P and $B equals pi over 4$ then $tan space A tan space B tan C$ =

In $capital delta$ ABC, D,E,F are the midpoints of sides BC, CA and AB respectively then ar $open parentheses B D E F close parentheses equals$

In $capital delta$ ABC, D,E,F are the midpoints of sides BC, CA and AB respectively then ar $open parentheses B D E F close parentheses equals$

If $tan space open parentheses x plus pi over 4 close parentheses equals a$ then $sec squared space x$ =

If $tan space open parentheses x plus pi over 4 close parentheses equals a$ then $sec squared space x$ =

$sin space pi over 7 sin space fraction numerator 2 pi over denominator 7 end fraction sin space fraction numerator 4 pi over denominator 7 end fraction$ =

$sin space pi over 7 sin space fraction numerator 2 pi over denominator 7 end fraction sin space fraction numerator 4 pi over denominator 7 end fraction$ =

Element $M plus N subscript 2 rightwards arrow with ΔH subscript 2 straight O on top NH subscript 3$ Element M belonging to group 13 can be

Element $M plus N subscript 2 rightwards arrow with ΔH subscript 2 straight O on top NH subscript 3$ Element M belonging to group 13 can be

In $blank subscript straight triangle ABC$ $A equals pi over 4 text and end text tan text end text B tan space C equals P n$ $t h e n space a l l space p o s s i b l e space v a l u e s space o f space P space i s$

In $blank subscript straight triangle ABC$ $A equals pi over 4 text and end text tan text end text B tan space C equals P n$ $t h e n space a l l space p o s s i b l e space v a l u e s space o f space P space i s$

In the given figure, ABCD is a Parallelogram then ar ( $capital delta$ AFB) is

In the given figure, ABCD is a Parallelogram then ar ( $capital delta$ AFB) is

In the following figure, if $capital delta$ ABC is circumscribed over the circle, then x =

In the following figure, if $capital delta$ ABC is circumscribed over the circle, then x =