Question

Answer the question using trigonometrical identities.​

Answer 1

TO PROVE :-

$( \sin \theta + \cos\theta)( \sec\theta + \csc\theta) = 2 + \sec\theta \csc\theta$

PROOF :-

LHS =

$( \sin \theta + \cos\theta)( \sec\theta + \csc\theta) \\ \\ \\ \implies ( \sin \theta + \cos\theta)( \frac{1}{ \cos \theta } + \frac{1}{ \sin \theta} ) \\ \\ \\ \implies \frac{\sin \theta}{ \cos \theta } + \frac{\sin \theta}{\sin \theta} + \frac{\cos \theta}{ \cos \theta} + \frac{ \cos \theta}{ \sin \theta} \\ \\ \\ \implies \: 2 + \frac{\sin \theta}{ \cos \theta } + \frac{ \cos \theta}{ \sin \theta} \\ \\ \implies 2 + \frac{ { \sin \theta}^{2} + { \cos \theta}^{2} }{ \cos \theta \sin \theta } \\ \\ \\ \implies 2 + \frac{1}{\cos \theta \sin \theta} \\ \\ \\ \boxed{ \implies 2 + \sec \theta \csc \theta}$

= RHS

Hence, Proved

SOME IDENTITIES :-

sin^2 + cos^2 = 1

sec^2 = tan^2 + 1

cosec^2 = sec^2 + 1

1/sinA = cosecA

1/tanA = cotA

1/cosA = secA

Answer 2

Well, first of all, you must read the chapter once and then come back over here which will be a great idea.

Here, what we all have to do is to change the trigononetric ratios "sin theta" and "cos theta" in terms of cosec theta and sec theta which we already learnt how to do.

The rest is all about how are you proceeding the solution and how carefulky are you observing the steps. If step is going all right, the answer will have to come correct as it is obtained above.

Several processes (or procedures) are there for solving these types of questions. You should accept the easier one.