Question

If I = m secA + n tanA and k = n secA + m tanA, then
(1) l² + k² = m² + n²
(2) l² - k²= m² + n²
(3) l² – k² = m² - n²
(4) l² + k² = m² - n²​

Answer 1

We know that ,

=>

${ \sec( \alpha ) }^{2} - { \tan( \alpha ) }^{2} = 1$

And ,

=> l = m secA + n tanA

squaring both sides ,

=> l² = m² sec²A + n² tan²A + 2mn secA tanA .......(1)

=> k = n secA + m tanA

squaring both sides ,

=> k² = n² sec²A + m² tan²A + 2mn secA tanA ........(2)

=> NOW , (1) - (2) ,

=> l² - k² = m² ( sec²A - tan²A ) + n² (-sec²A + tan²A ) + 2mn secA tanA - 2mn secA tanA

=> l² - k² = m² (1) + n² (-1)

=> l² - k² = m² - n²

So , the answer is option (3)!