Question

Prove( sins+secA)^2+(cosa+cosecA)^2=(1+secAcosecA)^2

Answer 1

I'll try to simplify:  

(sinA + secA)² + (cosA + cscA)²  

= (sinA + 1/cosA)² + (cosA + 1/sinA)²  

= sin²A + 2sinA/cosA + 1/cos²A + cos²A + 2cosA/sinA + 1/sin²A  

= (sin²A + cos²A) + (2sinA/cosA + 2cosA/sinA) + (1/cos²A + 1/sin²A)  

= 1 + 2(sin²A+cos²A)/(sinAcosA) + (sin²A+cos²A)/(sin²Acos²A)  

= 1 + 2/(sinAcosA) + 1/(sin²Acos²A)  

= (1 + 1/(sinAcosA))²  

= (1 + secA cscA)²  

This doesn't seem to match your solution since you have + sign after secA  

We can always go back 1 step and rewrite this differently:  

= (1 + 1/(sinAcosA))²  

= (1 + (sin²A+cos²A)/(sinAcosA))²  

= (1 + sin²A/(sinAcosA) + cos²A/(sinAcosA))²  

= (1 + sinA/cosA + cosA/sinA)²  

= (1 + tanA + cosA)²

Answer 2

aNS is given in pg 236 of rd sharma and i am sending u in whatsapp ok