Question

one upon sec a + 10 minus one upon Cos A it has to be proved that one upon Cos A minus one upon sec a minus 10

Answer 1

hope this answer helps.its just solved step by step and easy to understand...

Answer 2

Answer:

$\frac{1}{secA+tanA}-\frac{1}{cosA}=\frac{1}{cosA}-\frac{1}{secA-tanA}$

Step-by-step explanation:

$LHS=\frac{1}{secA+tanA}-\frac{1}{cosA}\\=\frac{(secA-tanA)}{(secA+tanA)(secA-tanA)}-secA\\=\frac{(secA-tanA)}{sec^{2}A-tan^{2}A)}-secA\\=\frac{(secA-tanA)}{1}-secA$

/* By Trigonometric identity:

sec²A-tan²A = 1*/

$=secA-tanA-secA\\=secA-(tanA+secA)\\=secA-(secA+tanA)\\=secA-\frac{(secA+tanA)(secA-tanA)}{(secA-tanA)}$

$=secA-\frac{sec^{2}A-tan^{2}A}{secA-tanA}$

$=secA-\frac{1}{secA-tanA}\\=\frac{1}{cosA}-\frac{1}{secA-tanA}\\=RHS$

Therefore,

$\frac{1}{secA+tanA}-\frac{1}{cosA}=\frac{1}{cosA}-\frac{1}{secA-tanA}$

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