Live practice

Use \(\sin^2A+\cos^2A=1\), \(1+\tan^2A=\sec^2A\), and \(1+\cot^2A=\cosec^2A\). The essential simplifications are:

1. (\(\cosec A-\cot A\)^2=[\(1-\cos A\)/\sin A]^2=\(1-\cos A\)/\(1+\cos A\)).
2. \[\begin{aligned}\dfrac{\cos A}{1+\sin A}+\dfrac{1+\sin A}{\cos A} &= \dfrac{2}{\cos A}\\&= 2\sec A\end{aligned}\].
3. After writing \(\tan A=\sin A/\cos A\) and \(\cot A=\cos A/\sin A\), the sum reduces to (1+1/\(\sin A\cos A\)=1+\sec A\cosec A).
4. (\(1+\sec A\)/\sec A=1+\cos A=\sin^2A/\(1-\cos A\)).
5. Multiplying numerator and denominator by \(\cos A+\sin A+1\) reduces the fraction to (\(1+\cos A\)/\sin A=\cosec A+\cot A).
6. (\sqrt{\(1+\sin A\)/\(1-\sin A\)}=\(1+\sin A\)/\cos A=\sec A+\tan A).
7. (\(\sin A-2\sin^3A\)/\(2\cos^3A-\cos A\)=\sin A\cos2A/\(\cos A\cos2A\)=\tan A).
8. Expanding the two squares and using the reciprocal identities gives \(7+\tan^2A+\cot^2A\).
9. (\(\cosec A-\sin A\)\(\sec A-\cos A\)=\sin A\cos A=1/\(\tan A+\cot A\)).
10. Both (\(1+\tan^2A\)/\(1+\cot^2A\)) and ([\(1-\tan A\)/\(1-\cot A\)]^2) simplify to \(\tan^2A\).

Formula or theorem used: Use \(PQ=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\) or the section formula \(\left(\frac{mx_2+nx_1}{m+n},\frac{my_2+ny_1}{m+n}\right)\).

Working: Substitute the coordinates before simplifying. For a shape, compare the required side lengths or diagonals and state the conclusion.

After substitution and simplification:

\(\left(-1,\frac72\right), (0,5), \left(1,\frac{13}{2}\right)\)

From \(\triangle ABE\cong\triangle ACD\), corresponding sides are equal:

Thus,

Also, \(D\) lies on \(AB\) and \(E\) lies on \(AC\), so the included angle is common:

By \(1\), \(2\), and SAS similarity, \(\triangle ADE\sim\triangle ABC\).

Formula or theorem used: For an AP use \(a_n=a+(n-1)d\) and \(S_n=\frac n2[2a+(n-1)d]\).

Working: Identify \(a\), \(d\), and the required term or sum; substitute them in the appropriate formula and simplify line by line.

After substitution and simplification:

13

For a uniform landing point, probability equals the ratio of areas. The rectangle area is \(3\times2=6\text{ m}^2\). The circle radius is \(0.5\) m, so its area is (\pi\(0.5\)^2=\pi/4).

Formula or theorem used: For grouped data use class marks and frequencies systematically. Apply the direct, assumed-mean, or step-deviation formula and show the working table totals.

Working: Write the relevant mean, median, or mode formula, identify the modal/median class, substitute the table totals, and interpret the numerical result in the question's context.

After substitution and simplification:

Median life = 3406.98 hours

The diagonal lengths are

Hence,

\(i\) \[\begin{aligned}\angle AEP &= \angle CDP\\&= 90^\circ\end{aligned}\], and \(\angle APE=\angle CPD\) are vertically opposite. Hence \(\triangle AEP\sim\triangle CDP\).

\(ii\) \[\begin{aligned}\angle ADB &= \angle CEB\\&= 90^\circ\end{aligned}\], and \(\angle ABD=\angle CBE\) because \(BD\) lies on \(BC\) and \(BE\) lies on \(BA\). Hence \(\triangle ABD\sim\triangle CBE\).

\(iii\) \[\begin{aligned}\angle AEP &= \angle ADB\\&= 90^\circ\end{aligned}\], and \(\angle EAP=\angle DAB\) is the angle between \(AB\) and \(AD\). Hence \(\triangle AEP\sim\triangle ADB\).

\(iv\) \[\begin{aligned}\angle PDC &= \angle BEC\\&= 90^\circ\end{aligned}\], and \(\angle PCD=\angle BCE\) is the angle between \(CE\) and \(CB\). Hence \(\triangle PDC\sim\triangle BEC\).

Formula or theorem used: For an AP use \(a_n=a+(n-1)d\) and \(S_n=\frac n2[2a+(n-1)d]\).

Working: Identify \(a\), \(d\), and the required term or sum; substitute them in the appropriate formula and simplify line by line.

After substitution and simplification:

4, 10, 16, 22, . . .

Good pens (=144-20=124).