Question

plz help me bro and sis

Answer 1

Hope it helps.......................

Answer 2

Step-by-step explanation:

In ΔABC,

cotθ = 12/5

       = AB/BC

Let AB = 12k, BC = 5k.

Now, AC = √AB² + BC²

               = √(12k)² + (5k)²

               = √144k² + 25k²

               = √169k²

               = 13k

So, AC = 13 k.

Thus, Sinθ = 5k/13k = 5/13.

         Cosθ = 12k/13k = 12/13.

         Secθ = 1/cosθ = 13/12

         Tanθ = sinθ/Cosθ = 5/12

         

LHS:

tan²θ - sin²θ

= (5/12)² - (5/13)²

= 25/144 - 25/169

= 625/24336

RHS:

sin⁴θ sec²θ

= (5/13)⁴ * (13/12)²

= 625/24336

∴ LHS = RHS

Hope it helps you!