Question

1+tan O / sin O + 1+cot 0/cos O =2(sec O + cosec 0)​

Answer 1

Step-by-step explanation:

LHS =1+cotA−cosecA=1+

sinA

cosA

−

sinA

1

=

sinA

(sinA+cosA−1)

1+tanA+secA=1+

cosA

sinA

+

cosA

1

=

cosA

(sinA+cosA+1)

(1+cotA−cosecA)(1+tanA+secA)

=

sinAcosA

(sinA+cosA−1)(sinA+cosA+1)

=

sinAcosA

(sinA+cosA)

2

−1

=

sinAcosA

1+2sinAcosA−1

=2=RHS

Hence Proved

Answer 2

Answer:

$lhs = \frac{1 + \tan \O }{ \sin \O } + \frac{1 + \cot \O }{ \sin \O } \\ \\ = \frac{1 + \frac{\sin \O}{cos\O\: } }{ \sin \O } + \frac{1 + \ \frac{cos \O}{sin \O} }{ \cos\O } \\ \\ = \frac{sin \O(cos \O + sin \O)}{cos \O} + \frac{cos \O(sin \O + cos\O)}{sin \O} \\ \\ = \frac{ {sin}^{2}\O (cos \O + sin \O) + {cos}^{2}\O( sin \O + cos\O)}{sin \O cos\O} \\ \\ = \frac{({sin}^{2}\O + {cos}^{2}\O)( sin \O + cos\O)}{sin \O cos\O} \\ \\ = \frac{sin \O}{sin \O cos\O} + \frac{cos\O}{sin \O cos\O} \\ \\ = \frac{1}{ cos\O} + \frac{1}{sin \O} \\ \\ = sec\O \: + cosec\O \\ = rhs$