Question

Consider a triangle PQR, right angled at P, in which PQ = 29 units, QR = 21 units and ∠PQR = θ, then find the values of
(i) cos²θ + sin²θ and (ii) cos²θ − sin²θ

Answer 1

here in question a small mistake is happened . ∆PQR is right angled at R not at P if PQ = 29{hypotenuse}

a triangle PQR, right angled at R, in which PQ = 29 units, QR = 21 units and ∠PQR = θ.
so, PR = √{29² - 21²}
= √{(29-21)(29+21)}
= √{8 × 50}
= 20 unit

(i) cos²θ + sin²θ
from right angled triangle PQR ,
cosθ = QR/PQ = 21/29
sinθ = PR/PQ = 20/29

now, sin²θ = 400/841
cos²θ = 441/841
so, sin²θ + cos²θ = 400/841 + 441/841
= (400 + 441)/841 = 841/841 = 1

(ii) cos²θ - sin²θ
= (21/29)² - (20/29)²
= 441/841 - 400/841
= (441 - 400)/841
= 41/841

Answer 2

I think it may be like this ,

In ∆PQR , <R = 90°

PQ = 29 units ,

QR = 21 units ;

RP² = PQ² - QR²

[ By Phythogarian theorem ]

RP² = 29² - 21²

= ( 29 + 21 )( 29 - 21 )

= 50 × 8

= 400

RP =√400

RP = 20 units

i ) sin² theta + cos² theta

= ( RP/PQ )² + ( QR/PQ )²

= ( RP² + QR² )/PQ²

= PQ²/PQ²

= 1

ii ) cos² theta - sin² theta

= ( QR/PQ )² - ( RP/PQ )²

= ( 21/29 )² - ( 20/29 )²

= 21²/29² - 20²/29²

= ( 21² - 20² )/29²

= [( 21 + 20 ) ( 21 - 20 )]/29²

= ( 41 × 1 )/841

= 41/841

I hope this helps you.

: )