Question
Evaluate the following:
$\text{cosec 31}^\circ-\sec59^\circ$
$\text{cosec 31}^\circ-\sec59^\circ$
✓
Answer
We have to find: $\text{cosec 31}^\circ-\sec59^\circ$$$Since $\text{cosec}(90^\circ-\theta)=\sec\theta$
So, $=\text{cosec 31}^\circ-\sec59^\circ$ $=\text{cosec}(90^\circ-59^\circ)-\sec59^\circ$ $\sec59^\circ-\sec59^\circ$ $= 0$So value of $=\text{cosec 31}^\circ-\sec59^\circ\text{ is 0}$
So, $=\text{cosec 31}^\circ-\sec59^\circ$ $=\text{cosec}(90^\circ-59^\circ)-\sec59^\circ$ $\sec59^\circ-\sec59^\circ$ $= 0$So value of $=\text{cosec 31}^\circ-\sec59^\circ\text{ is 0}$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Find the median of the following frequency distribution:
|
Marks
|
0-10
|
10-20
|
20-30
|
30-40
|
40-50
|
|
Number of students
|
6
|
16
|
30
|
9
|
4
|
In figure, chord EF || chord GH. Prove that, chord EG ≅ chord FH. Fill in the blanks and write the proof.
Proof: Draw seg GF.
$\angle EFG =\angle FGH \quad \ldots . . . \square \quad \ldots . . .( I )$
$\angle EFG =\square \quad \ldots . .[\text { [inscribed angle theorem] (II) }$
$\angle FGH =\square \quad \ldots . .[\text { inscribed angle theorem] (III) }$
$\therefore m (\operatorname{arc} EG )=\square \quad \ldots \ldots[ By ( I ),( II ), \text { and (III)] }$
chord $EG \cong$ chord $FH \quad$.........[corresponding chords of congruent arcs]
→Proof: Draw seg GF.
$\angle EFG =\angle FGH \quad \ldots . . . \square \quad \ldots . . .( I )$
$\angle EFG =\square \quad \ldots . .[\text { [inscribed angle theorem] (II) }$
$\angle FGH =\square \quad \ldots . .[\text { inscribed angle theorem] (III) }$
$\therefore m (\operatorname{arc} EG )=\square \quad \ldots \ldots[ By ( I ),( II ), \text { and (III)] }$
chord $EG \cong$ chord $FH \quad$.........[corresponding chords of congruent arcs]
In the given figure, LM || CD.
Prove that $\frac{\text{AM}}{\text{AB}}=\frac{\text{AN}}{\text{AD}}.$
→Prove that $\frac{\text{AM}}{\text{AB}}=\frac{\text{AN}}{\text{AD}}.$
Find the area of a circle whose circurnference is $8\pi.$
→Prove the following trigonometric identities.
$\frac{\cot\text{A}-\cos\text{A}}{\cot\text{A}+\cos\text{A}}=\frac{\text{cosec A}-1}{\text{cosec A}+1}$
→$\frac{\cot\text{A}-\cos\text{A}}{\cot\text{A}+\cos\text{A}}=\frac{\text{cosec A}-1}{\text{cosec A}+1}$
The graph of the polynomial $f(x) = ax^2 + bx + c$ is as shown below Fig. Write the signs of $'a'$ and $b^2 - 4ac.$
→In a family of 3 children, find the probability of having at least one boy.
The area of a sector of a circle of radius 2cm is $\pi\text{cm}^2.$ Find the angle contained by the sector.
→A Wire when bent in the form of an equilateral triangle encloses an area of $121\sqrt{3}\text{cm}^2.$ If the same wire is bent into the form of a circle, what will be the area of the circle? $\Big[\text{Take }\pi=\frac{22}{7}.\Big]$
→Evaluate the following:
$\frac{\sin30^\circ}{\cos45^\circ}+\frac{\cot45^\circ}{\sec60^\circ}-\frac{\sin60^\circ}{\tan45^\circ}+\frac{\cos30^\circ}{\sin90^\circ}$
→$\frac{\sin30^\circ}{\cos45^\circ}+\frac{\cot45^\circ}{\sec60^\circ}-\frac{\sin60^\circ}{\tan45^\circ}+\frac{\cos30^\circ}{\sin90^\circ}$