Question

Solve the problem 2....

Answer 1

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Answer 2

sin θ ( 1 + tan θ ) + cos θ ( 1 + cot θ ) 

sin θ ( 1 + sin θ / cos θ ) + cos θ ( 1 + cos θ / sin θ ) 

sin θ ( cos θ + sin θ ) / cos θ + cos θ ( sin θ + cos θ ) / sin θ 

( cos θ + sin θ ) ( tan θ + cot θ ) 

( cos θ + sin θ ) ( sin² θ + cos²θ ) / ( sin θ cos θ ) 

( cos θ + sin θ ) / ( sin θ cos θ ) since sin² θ + cos²θ = 1 

[ cos θ / ( sin θ cos θ ) ] + [ sin θ / ( sin θ cos θ ) ] 

1 / sin θ + 1 / cos θ 

cosec θ + sec θ = ( Hence Proved )