Question

10th CBSE Trigonometry ​

Answer 1

Answer:

Step-by-step explanation:

Answer 2

To prove :-

(1 + secA)/secA = sin²A/(1 - cosA)

$LHS : - \\ \\ \sf = \frac{1 + sec A}{secA} \\ \\ \sf = \frac{1 + \frac{1}{cosA} }{ \frac{1}{cosA} } \\ \\ \sf = \frac{cosA + 1}{cosA} \times \frac{cosA}{1} \\ \\ \sf = \frac{ {cos}^{2}A + cosA }{cosA} \\ \\ \sf = \frac{cosA(cosA + 1)}{cosA} \\ \\ \sf = cosA + 1 \\ \\ RHS :- \\ \\ \sf = \frac{ {sin}^{2} A}{1 - cosA} \\ \\ \sf = \frac{1 - {cos}^{2}A }{1 - cosA} \\ \\ \sf = \frac{( 1 - cosA)(1 + cosA)}{(1 - cosA)} \\ \\ \sf = 1 + cosA \\ \\ LHS = RHS$

HENCE PROVED!

Important things to note which are used here :-

• secA = 1/cosA

• cos²A + sin²A = 1

➡ sin²A = 1 - cos²A