Question

if tanA+sinA=m and tanA-sinA=n then prove that m²-n²=4√mn​

Answer 1

m²- n²  = 4 $\sqrt{mn}$   proved

Step-by-step explanation:

Given as :

The trigonometrical function

tan A + sin A = m                  ........1

tan A - sin A = n                   ..........2

To prove :  m²- n²  = 4 $\sqrt{mn}$

According to question

From left hand side

put the value of m and n

i.e   m² - n²  =   ( tan A + sin A )²- ( tan A - sin A )² = [ ( tan² A + sin² A + 2 tan A sin A ) - (     tan² A + sin² A - 2 tan A sin A ) ]

                                                             = [ tan² A + sin² A + 2 tan A sin A - tan² A - sin² A + 2 tan A sin A ) ]

                                                            = [ ( tan² A -  tan² A ) + ( Sin² A -  Sin² A ) + 4 tan A sin A

                                                            = [ 0 + 0 + 4 tan A sin A ]

                                                            = 4 tan A sin A

So,  m² - n² = 4 tan A sin A                      ...........3

From right hand side

i.e  4 $\sqrt{mn}$ = 4 × $\sqrt{(tan A + sin A) ( tan A - sin A )}$

                 =  4 × $\sqrt{tan^{2} A - sin^{2}A }$

                 = 4 $\sqrt{sin^{2} A ( \dfrac{1-cos^{2}A }{cos^{2} A}) }$

                 = 4 $\sqrt{\dfrac{sin^{4} A}{cos^{2} A} }$

                  = 4 $\dfrac{sin^{2}A }{cosA}$

                  = 4 sin A × $\dfrac{sinA}{cosA}$

                  = 4 tan A sin A

So, 4 $\sqrt{mn}$   =  = 4 tan A sin A                            ............4

Now, From eq 3 and eq 4 ,

Left hand side = Right hand side

So, m²- n²  = 4 $\sqrt{mn}$

Hence, m²- n²  = 4 $\sqrt{mn}$   proved  . Answer

Answer 2

Answer:

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