Question

1/secQ+tanQ =1-sinQ/cosQ​

Answer 1

Answer:

Step-by-step explanation:

Multiply and divide with secQ-tanQ

We get

secQ-tanQ/sec^2Q-tan^2Q

=secQ-tanQ

=1/cosQ -sinQ/cosQ

=1-sinQ/cosQ

Answer 2

QUESTION:

$\frac{1}{ \sec(q) + \tan(q) } = \frac{1 - \sin(q) }{ \cos(q) }$

FORMULAS USED:

→(a+b)(a-b) = a²-b²

ANSWER:

In question taking LHS

multiply and divide with secQ - tanQ

$\frac{1}{ \sec(q) + \tan(q) } \times \frac{ \sec(q) - \tan(q) }{ \sec(q) - \tan(q) } \\ \\ = \frac{ \sec(q) - \tan(q) }{ { \sec(q) }^{2} - { \tan(q) }^{2} } \\ \\ = \sec(q) - \tan(q) \\ \\ = \frac{1}{ \cos(q) } - \frac{ \sin(q) }{ \cos(q) } \\ \\ = \frac{1 - \sin(q) }{ \cos(q) }$

LHS=RHS