Question

secA-1/secA+1=1-cosA/1+cos​

Answer 1

secA-1/secA+1=1-cosA/1+cos

LHS

=(1/cosA - 1)/ (1/cosA+1)

=(1-cosA/cosA)/(1+cosA/cosA)

=1-cosA/1+cosA

Step-by-step explanation:

secA= 1/cosA

Answer 2

Given :

• $\large \red{ \sf{\dfrac{secA-1}{secA+1}}}$

To Prove :

• $\large\orange{\sf{\dfrac{secA-1}{secA+1}} = \dfrac{1 - cosA}{1 + cosA}}$

Solution :

$\large\sf{LHS \: = \blue {\dfrac{secA - 1}{secA + 1}}} \\ \\ \\ \implies\large \purple{ \sf{ \dfrac{ \dfrac{1}{cos A} - 1 }{ \dfrac{1}{cos A} + 1 }}} \\ \\ \\ \implies\large\red{\sf{ \dfrac{1 - cosA}{1 + cosA}}} \\ \\ \\ \sf {= \: RHS}$

Addition information :

• tanø = sinø/cosø

• secø = 1/cosø

• cotø = 1/tanø = sinø/cosø

• 1 - tan(ø/2)/1 - tan(ø/2) = ±√1 - sinø/1 + sinø

• tan ø/2 = ±√1 - cosø/1 + cosø

• sinø = Cos(90° - ø)

• cosø= sin(90° - ø)

• tanø = cot(90° - ø)

• cotø = tan(90° - ø)

• secø = cosec(90° - ø)