ICSE Guess > ICSE/ISC eBooks > Class X > Maths by Mr. M. P. Keshari

ICSE / ISC eBooks

Example – 3. Prove that : (cosecA – sinA).(secA – cosA)  =  1/(tanA + cotA)

Solution :-   LHS  =  (cosecA – sinA).(secA – cosA)

=  (1/sinA – sinA).(1/cosA – cosA)

=  (1 – sin2A)/sinA.(1 – cos2A)/cosA

=  (cos2A/sinA).(sin2A/cosA)

=  sinA.cosA  …….. (i)

RHS  =  1/(tanA + cotA) 

=  1/(sinA/cosA + cosA/sinA)

=   1/{(sin²A + cos²A)/cosAsinA}   [sin²A + cos²A = 1]

= sinA.cosA  ………. (ii)

From (i)  and  (ii)  we get    LHS  =  RHS.

Example – 4. If tanA + sinA = m and tanA – sinA = n, show that m² – n² = 4√(mn)

Solution :- LHS = m² – n² = (tanA + sinA)² – (tanA – sinA)²
                           = tan²A + sin²A + 2tanAsinA – tan²A – sin²A + 2tanAsinA
                           = 4tanA.sinA 
                   RHS =  4√(mn)
                          = 4√(tanA + sinA).(tanA – sinA )
                          = 4√(tan²A – sin²A

    = 4√(sin²A/cos²A – sin²A)
                            = 4√{(sin²A – sin²Acos²A)/cos²A}
                            = 4√{sin²A(1 – cos²A)/cos²A}
                            = 4√(sin²A.sin²A/cos²A)
                            = 4(sinA.sinA/cosA)
                            = 4.tanA.sinA  =  LHS