Question

If 7secA=25, then the value of cosec²A—cot²A is______​

Answer 1

Given:

• 7 secA = 25

Find:

• cosec² A - cot² A

Solution:

⇒ 7 secA = 25

⇒ sec A $\rm =\dfrac{25}{7} = \dfrac{H}{B}$

So,

• Hypotenuse, H = 25
• Base, B= 7

So,

$\setlength{\unitlength}{1cm}\begin{picture}(6,5)\linethickness{0.5mm}\put(1,1){\line(1,0){4.5}}\put(1,1){\line(0,1){3.5}}\qbezier(1,4.5)(1,4.5)(5.5,1){\framebox(0.3,0.3)}\put(0.7,4.6){\large\bf C}\put(.8,0.6){\large\bf B}\put(5.6,0.6){\large\bf A}\qbezier(4.5,1)(4.3,1.25)(4.6,1.7)\put(4,1.4){ \bf\Theta } \put(3,0.5){ \bf 7} \put(3,3){ \bf 25} \put(1.4,1){\line(0,1){0.4}} \put(1,1.4){\line(1,0){0.4}} \end{picture}$

Now,

we, know that

$\boxed{\rm H^2 = P^2 + B^2}$

where,

• H = 25
• B = 7

So,

$:\to\rm H^2 = P^2 + B^2$

$:\to\rm (25)^2 = P^2 + (7)^2$

$:\to\rm 625= P^2 + 49$

$:\to\rm 625 - 49= P^2$

$:\to\rm 576= P^2$

$:\to\rm \sqrt{576}= P$

$:\to\rm 24= P$

$:\to\rm P = 24$

$\therefore\rm P = 24$

_________________________

Now,

$\to \rm \cosec A = \dfrac{H}{P}$

where,

• H = 25
• P = 24

So,

$\to \rm \cosec A = \dfrac{H}{P}$

$\to \rm \cosec A = \dfrac{25}{24} .......1$

Now,

$\to \rm \cot A = \dfrac{ B}{P}$

where,

• B = 7
• P = 24

So,

$\to \rm \cot A = \dfrac{ B}{P}$

$\to \rm \cot A = \dfrac{7}{24} ........2$

we, have to find value of

$\rm \dashrightarrow { \cosec}^{2} A - { \cot}^{2} A$

Using eq(1) and eq(2) here, we get

$\rm \dashrightarrow { \bigg(\dfrac{25}{24} \bigg) }^{2} - { \bigg(\dfrac{7}{24} \bigg)}^{2}$

$\rm \dashrightarrow \dfrac{625}{576}- \dfrac{49}{576}$

$\rm \dashrightarrow \dfrac{576}{576}$

$\rm \dashrightarrow 1$

_________________________

$\rm \therefore { \cosec}^{2} A - { \cot}^{2} A = 1$