Question

(a) Prove that:
$(1+tanA{)}^2+(1-tanA{)}^2=2se{c}^{2}A$

(b) If sin (A + B) = 1 and cos (A - B) = 1, find the values of A and B. Given $0\le {90}^{\circ }$ and $A\ge B$ [6 MARKS]

Answer 1

Each Part: 3 Marks

(a)LHS $=(1+tanA{)}^2+(1-tanA{)}^2$

$={(1+\frac{sinA}{cosA})}^2+{(1-\frac{sinA}{cosA})}^2$

$=\frac{(cosA+sinA{)}^2}{co{s}^{2}A}+\frac{(cosA-sinA{)}^2}{co{s}^{2}A}$

$=\frac{(co{s}^{2}A+si{n}^{2}A+2sinAcosA)+(co{s}^{2}A+si{n}^{2}A-2sinAcosA)}{co{s}^{2}A}$

$=\frac{2si{n}^{2}A+2co{s}^{2}A}{co{s}^{2}A}=\frac{2(si{n}^{2}A+co{s}^{2}A)}{co{s}^{2}A}$

$=\frac{2}{co{s}^{2}A}=2se{c}^{2}A=RHS$

Hence, proved

(b) Given: $sin(A+B)=1$
$\Rightarrow sin(A+B)=sin{90}^{\circ }$ $[\because sin{90}^{\circ }=1]$
$\Rightarrow sin(A+B)=sin{90}^{\circ }\dots \dots \left(i\right)$
Also $cos(A-B)=1$
$\Rightarrow cos(A-B)=cos{0}^{\circ }$ $[\because cos{0}^{\circ }=1]$
$\Rightarrow A-B=0\dots \dots \left(ii\right)$
Solving equations (i) and (ii) we get; $A={45}^{\circ }$ and $B={45}^{\circ }$
Hence, $A=B={45}^{\circ }$