Question

sinQ - cosQ + 1 / sinQ + cos Q - 1 = 1/secQ - tanQ

Answer 1

$\boxed{\underline{\underline{\textsf{Formulas :}}}}$

$1.\:sinQ=2sin\frac{Q}{2}cos\frac{Q}{2}$

$2.\:cosQ=2cos^{2}\frac{Q}{2}-1$

$3.\:cosQ=1-2sin^{2}\frac{Q}{2}$

$4.\:sin^{2}\frac{Q}{2}+cos^{2}\frac{Q}{2}=1$

$5.\:a^{2}-b^{2}=(a+b)(a-b)$

$6.\:(a+b)^{2}=a^{2}+2ab+b^{2}$

$7.\:sec^{2}Q-tan^{2}Q=1$

$\boxed{\underline{\underline{\textsf{Proof 1 - ( For Class 10th ) :}}}}$

$\textsf{Now,}\:\frac{sinQ-cosQ+1}{sinQ+cosQ-1}$

$=\frac{tanQ-1+secQ}{tanQ+1-secQ}$

$\textsf{divided both the numerator and}$
$\textsf{the denominator by cosQ}$

$=\frac{(tanQ+secQ-1)}{(tanQ-secQ)+1}$

$=\frac{(tanQ+secQ-1)(tanQ+secQ)}{\{(tanQ-secQ)+1\}(tanQ+secQ)}$

$\textsf{multiplied both the numerator and}$
$\textsf{the denominator by (tanQ + secQ)}$

$=\frac{(tanQ+secQ-1)(tanQ+secQ)}{(tanQ-secQ)(tanQ+secQ)+(tanQ+secQ)}$

$=\frac{(tanQ+secQ-1)(tanQ+secQ)}{tan^{2}Q-sec^{2}Q+tanQ+secQ}$

$=\frac{(tanQ+secQ-1)(tanQ+secQ)}{-1+tanQ+secQ}$

$=tanQ+secQ$

$=\frac{1}{secQ-tanQ}$

$\to \boxed{\small{\frac{sinQ-cosQ+1}{sinQ+cosQ-1}=\dfrac{1}{secQ-tanQ}}}$

$\underline{\textsf{Hence, proved.}}$

$\boxed{\underline{\underline{\textsf{Proof 2 - ( Higher Classes ) :}}}}$

$Now,\:\frac{sinQ-cosQ+1}{sinQ+cosQ-1}$

$=\frac{sinQ+(1 -cosQ)}{sinQ-(1-cosQ)}$

$=\dfrac{2sin\frac{Q}{2}cos\frac{Q}{2}+2sin^{2}\frac{Q}{2}}{2sin\frac{Q}{2}cos\frac{Q}{2}-2sin^{2}\frac{Q}{2}}$

$=\dfrac{sin\frac{Q}{2}(cos\frac{Q}{2}+sin\frac{Q}{2})}{sin\frac{Q}{2}(cos\frac{Q}{2}-sin\frac{Q}{2})}$

$=\dfrac{(cos\frac{Q}{2}+sin\frac{Q}{2})(cos\frac{Q}{2}+sin\frac{Q}{2})}{(cos\frac{Q}{2}-sin\frac{Q}{2})(cos\frac{Q}{2}+sin\frac{Q}{2})}$

$=\dfrac{cos^{2}\frac{Q}{2}+sin^{2}\frac{Q}{2}+2sin\frac{Q}{2}cos\frac{Q}{2}}{cos^{2}\frac{Q}{2}-sin^{2}\frac{Q}{2}}$

$=\dfrac{1+sinQ}{cosQ}$

$=\dfrac{1}{cosQ}+\dfrac{sinQ}{cosQ}$

$=secQ+tanQ$

$=\dfrac{1}{secQ-tanQ}$

$\to \boxed{\small{\frac{sinQ-cosQ+1}{sinQ+cosQ-1}=\dfrac{1}{secQ-tanQ}}}$

$\underline{\textsf{Hence, proved.}}$

$\boxed{\underline{\underline{\textsf{Proof 3 - ( Long Method ) :}}}}$

$Now,\:\frac{sinQ-cosQ+1}{sinQ+cosQ-1}$

$=\dfrac{2sin\frac{Q}{2}cos\frac{Q}{2}-(2cos^{2}\frac{Q}{2}-1)+1}{2sin\frac{Q}{2}cos\frac{Q}{2}+(2cos^{2}\frac{Q}{2}-1)-1}$

$=\dfrac{2sin\frac{Q}{2}cos\frac{Q}{2}-2cos^{2}\frac{Q}{2}+1+1}{2sin\frac{Q}{2}cos\frac{Q}{2}+2cos^{2}\frac{Q}{2}-1-1}$

$=\dfrac{2sin\frac{Q}{2}cos\frac{Q}{2}-2cos^{2}\frac{Q}{2}+2}{2sin\frac{Q}{2}cos\frac{Q}{2}+2cos^{2}\frac{Q}{2}-2}$

$=\dfrac{2(sin\frac{Q}{2}cos\frac{Q}{2}-cos^{2}\frac{Q}{2}+1)}{2(sin\frac{Q}{2}cos\frac{Q}{2}+cos^{2}\frac{Q}{2}-1)}$

$=\small{\dfrac{sin\frac{Q}{2}cos\frac{Q}{2}-cos^{2}\frac{Q}{2}+sin^{2}\frac{Q}{2}+cos^{2}\frac{Q}{2}}{sin\frac{Q}{2}cos\frac{Q}{2}+cos^{2}\frac{Q}{2}-sin^{2}\frac{Q}{2}+cos^{2}\frac{Q}{2}}}$

$=\dfrac{sin\frac{Q}{2}cos\frac{Q}{2}+sin^{2}\frac{Q}{2}}{sin\frac{Q}{2}cos\frac{Q}{2}-sin^{2}\frac{Q}{2}}$

$=\dfrac{sin\frac{Q}{2}(cos\frac{Q}{2}+sin\frac{Q}{2})}{sin\frac{Q}{2}(cos\frac{Q}{2}-sin\frac{Q}{2})}$

$=\dfrac{(cos\frac{Q}{2}+sin\frac{Q}{2})(cos\frac{Q}{2}+sin\frac{Q}{2})}{(cos\frac{Q}{2}-sin\frac{Q}{2})(cos\frac{Q}{2}+sin\frac{Q}{2})}$

$=\dfrac{cos^{2}\frac{Q}{2}+sin^{2}\frac{Q}{2}+2sin\frac{Q}{2}cos\frac{Q}{2}}{cos^{2}\frac{Q}{2}-sin^{2}\frac{Q}{2}}$

$=\dfrac{1+sinQ}{cosQ}$

$=\dfrac{1}{cosQ}+\dfrac{sinQ}{cosQ}$

$=secQ+tanQ$

$=\dfrac{1}{secQ-tanQ}$

$\to \boxed{\small{\frac{sinQ-cosQ+1}{sinQ+cosQ-1}=\dfrac{1}{secQ-tanQ}}}$

$\underline{\textsf{Hence, proved.}}$

$\boxed{\underline{\underline{\textsf{Finding (secQ + tanQ)}}}}$

$\textsf{We know that,}$

$sec^{2}Q-tan^{2}Q=1$

$\to \small{(secQ+tanQ)(secQ-tanQ)=1}$

$\to secQ+tanQ=\dfrac{1}{secQ-tanQ}$

Answer 2

Step-by-step explanation:

SinQ - CosQ + 1 / SinQ + CosQ - 1 = 1/SecQ - TanQ

LHS = SinQ/CosQ - CosQ/CosQ + 1/CosQ / SinQ/CosQ + CosQ/CosQ - 1/Cos . ( dividing the whole LHS by CosQ)

= (TanQ + SecQ) - 1 / (TanQ - SecQ) + 1

= {(TanQ + SecQ) - 1} (TanQ - SecQ) / {(TanQ - SecQ) + 1} ( TanQ - SecQ). {Multiplying the numerator and denominator by (TanQ - SecQ)}

= (Tan^2Q - Sec^2Q) - ( TanQ - SecQ) / (TanQ - SecQ + 1) ( TanQ - SecQ). ( Here I have just multiplied numerator wholly)

= -1 - (TanQ - SecQ) / (TanQ - SecQ) (TanQ - SecQ + 1)

= -1/TanQ - SecQ = 1/SecQ - TanQ . { by cancelling (TanQ - SecQ) we get this }

LHS = RHS ( Hence Proved)

Hope it helps!☺

Have a nice day!❤