Question

If 3cotA=4 then find Sin²A - Sec²A

Answer 1

here you go ☆☆

$\cot(a) = \frac{4}{3}$
so,
given:-
◇base = 4
◇perpendicular = 3

to find :-
◇h = ?
◇Sin²A - Sec²A

using Pythagoras theorem

▪${h }^{2} = {b}^{2} + {p}^{2}$
▪${h}^{2} = 16 + 9 = 25$
▪$h = \sqrt{25} = > 5$

□sin A = 3/5

□sec A =5/4

▪${ \sin }^{2} - { \sec }^{2}$

▪${ \frac{3}{5} }^{2} - { \frac{5}{4} }^{2}$

▪$\frac{9}{25} - \frac{25}{16}$

▪$\frac{ 144 - 625}{400}$

▪$- \frac{481}{400}$

▪$1.20$

hope it helps:)

Answer 2

let b = base , p = perpendicular and
h = hypotenuse

3cotA = 4 => cotA = 4 / 3 = b/ p

solve for ' h' :
------------------
by Pythagoras theorem, we get

h^2 = ( b)^2 + ( p )^2

h^2 = ( 4 )^2 + ( 3 )^2

h^2 = 16 + 9 = 25

h = √25 => h = 5

Find the value of SinA :
--------------------------------

since, we know sinA = p / h

sinA = 3 / 5

Find the value of SecA :
---------------------------------

secA = h / b = 5 / 4

so now put the value of sinA and secA in the Given trigonometric expression :

sin^2A - sec^2A = ( 3 / 5 )^2 - ( 5 / 4 )^2

= ( 9 / 25 ) - ( 25 / 16 )

= ( 144 - 625 ) / 400

= - 481 / 400

therefore, sin^2A - sec^2A = - 481 / 400

Answer : - 481 / 400

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