Solutions to SL Loney Plane Trigonometry Chapter 2

Solutions to SL Loney Plane Trigonometry Chapter 2


Herein you will get Solutions to SL Loney Plane Trigonometry Chapter 2, Examples 5, 6 and 7. Remember, Loney calls Exercises by the name of Examples. By design, we are providing solutions to only select items leaving the rest for the much-needed practice at your end.

Jumpstart with the examples here. Or better, start over systematically with the first chapter i.e. Measurement of Angles. You may otherwise choose to go through the post on Trigonometrical Ratios of the Sum and Difference of Two Angles.

Trigonometrical Ratios (in terms of each-other)

\[
sin\theta
\]
\[
cos\theta
\]
\[
tan\theta
\]
\[
cot\theta
\]
\[
sec\theta
\]
\[
cosec\theta
\]
\[
sin\theta
\]
\[
sin\theta
\]
\[
\sqrt{1 – cos^2\theta}
\]
\[
\frac{tan\theta}{\sqrt{1 + tan^2\theta}}
\]
\[
\frac{1}{\sqrt{1 + cot^2\theta}}
\]
\[
\frac{\sqrt{sec^2\theta – 1}}{sec\theta}
\]
\[
\frac{1}{cosec\theta}
\]
\[
cos\theta
\]
\[
\sqrt{1 – sin^2\theta}
\]
\[
cos\theta
\]
\[
\frac{1}{\sqrt{1 + tan^2\theta}}
\]
\[
\frac{cot\theta}{\sqrt{1 + cot^2\theta}}
\]
\[
\frac{1}{sec\theta}
\]
\[
\frac{\sqrt{cosec^2\theta – 1}}{cosec\theta}
\]
\[
tan\theta
\]
\[
\frac{sin\theta}{\sqrt{1 – sin^2\theta}}
\]
\[
\frac{\sqrt{1 – cos^2\theta}}{cos\theta}
\]
\[
tan\theta
\]
\[
\frac{1}{cot\theta}
\]
\[
\sqrt{sec^2\theta – 1}
\]
\[
\frac{1}{\sqrt{cosec^2\theta – 1}}
\]
\[
cot\theta
\]
\[
\frac{\sqrt{1 – sin^2\theta}}{sin\theta}
\]
\[
\frac{cos\theta}{\sqrt{1 – cos^2\theta}}
\]
\[
\frac{1}{tan\theta}
\]
\[
cot\theta
\]
\[
\frac{1}{\sqrt{sec^2\theta – 1}}
\]
\[
\sqrt{cosec^2\theta – 1}
\]
\[
sec\theta
\]
\[
\frac{1}{\sqrt{1 – sin^2\theta}}
\]
\[
\frac{1}{cos\theta}
\]
\[
\sqrt{1 + tan^2\theta}
\]
\[
\frac{\sqrt{1 + cot^2\theta}}{cot\theta}
\]
\[
sec\theta
\]
\[
\frac{cosec\theta}{\sqrt{cosec^2\theta – 1}}
\]
\[
cosec\theta
\]
\[
\frac{1}{sin\theta}
\]
\[
\frac{1}{\sqrt{1 – cos^2\theta}}
\]
\[
\frac{\sqrt{1 + tan^2\theta}}{tan\theta}
\]
\[
\sqrt{1 + cot^2\theta}
\]
\[
\frac{sec\theta}{\sqrt{sec^2\theta – 1}}
\]
\[
cosec\theta
\]

SL Loney – Examples 5

Prove that

#12. \[
\frac{1 + tan^2A}{1 + cot^2A} = \frac{sin^2A}{cos^2A}
\]Solution: \[
LHS = \frac{sec^2A}{csc^2A} = \frac{1}{cos^2A} \div \frac{1}{sin^2A}
\]\[
= \frac{sin^2A}{cos^2A} = RHS
\]

#15. \[
\frac{cosA}{1 – tanA} + \frac{sinA}{1 – cotA} = sinA + cosA
\]Solution: \[
LHS = \frac{cosA}{1 – \frac{sinA}{cosA}} + \frac{sinA}{1 – \frac{cosA}{sinA}}
\]\[
= \frac{cosA}{\frac{cosA – sinA}{cosA}} + \frac{sinA}{\frac{sinA – cosA}{sinA}}
\]\[
= \frac{cos^2A}{cosA – sinA} + \frac{sin^2A}{sinA – cosA}
\]\[
= \frac{cos^2A}{cosA – sinA} – \frac{sin^2A}{cosA – sinA}
\]\[
= \frac{cos^2A – sin^2A}{cosA – sinA} = \frac{(cosA + sinA)(cosA – sinA)}{cosA – sinA}
\]\[
= cosA + sinA = RHS
\]

#17. \[
sec^4A – sec^2A = tan^4A + tan^2A
\]Solution: \[
LHS = (1 + tan^2A)^2 – (1 + tan^2A)
\]\[
= (1 + 2tan^2A + tan^4A) – 1 – tan^2A
\]\[
= tan^4A + tan^2A = RHS
\]

#19. \[
\sqrt{cosec^2A – 1} = cosA.cosecA
\]Solution: \[
LHS = \sqrt{1 + cot^2A – 1}
\]\[
= \sqrt{cot^2A} = cotA
\]\[
= \frac{cosA}{sinA} = cosA.cosecA = RHS
\]

Useful Link

#21. \[
tan^2A – sin^2A = sin^4A.sec^2A
\]Solution: \[
LHS = \frac{sin^2A}{cos^2A} – sin^2A
\]\[
= \frac{sin^2A – sin^2Acos^2A}{cos^2A} = \frac{sin^2A(1 – cos^2A)}{cos^2A}
\]\[
= \frac{sin^2A}{cos^2A} \times sin^2A = sin^4A \times \frac{1}{cos^2A}
\]\[
= sin^4A.sec^2A = RHS
\]

Alter: You could do this item as easily taking the RHS. \[
RHS = sin^2A \times sin^2A \times sec^2A
\]\[
= (1 – cos^2A) \times sin^2A \times \frac{1}{cos^2A}
\]\[
= (1 – cos^2A)tan^2A
\]\[
= tan^2A – cos^2A\frac{sin^2A}{cos^2A}
\]\[
= tan^2A – sin^2A = LHS
\]

#23. \[
\frac{1}{cosecA – cotA} – \frac{1}{sinA} = \frac{1}{sinA} – \frac{1}{cosecA + cotA}
\]Solution: Let’s take the two sides one by one. \[
RHS = \frac{1}{sinA} – \frac{1}{\frac{1}{sinA} + \frac{cosA}{sinA}}
\]\[
= \frac{1}{sinA} – \frac{1}{\frac{1 + cosA}{sinA}}
\]\[
= \frac{1}{sinA} – \frac{sinA}{1 + cosA}
\]\[
= \frac{1 + cosA – sin^2A}{sinA(1 + cosA)}
\]\[
= \frac{1 + cosA – (1 – cos^2A)}{sinA(1 + cosA)}
\]\[
= \frac{cosA + cos^2A}{sinA(1 + cosA)}
\]\[
= \frac{cosA(1 + cosA)}{sinA(1 + cosA)} = cotA
\]\[
LHS = \frac{cosecA + cotA}{(cosecA – cotA)(cosecA + cotA)} – cosecA
\]\[
= \frac{cosecA + cotA}{cosec^2A – cot^2A} – cosecA
\]\[
= \frac{cosecA + cotA}{1 + cot^2A – cot^2A} – cosecA
\]\[
= cosecA + cotA – cosecA = cotA = RHS
\]

#25. \[
\frac{cotA + tanB}{cotB + tanA} = cotA.tanB
\]Solution: \[
LHS = \frac{cotA + tanB}{\frac{1}{tanB} + \frac{1}{cotA}} = \frac{cotA + tanB}{\frac{cotA + tanB}{cotA.tanB}}
\]\[
= cotA.tanB = RHS
\]

SL Loney – Examples 6

#1. Express all the trigonometrical ratios in terms of \(cosine.\)

Hint: Since \(sin^2\theta+cos^2\theta=1\)

\(\Rightarrow sin\theta = \sqrt{1 – cos^2\theta}\)

Others ratios flow herefrom.

#2. Express all the trigonometrical ratios in terms of \(tangent.\)

Hint: Start with \(sec^2\theta = 1 + tan^2\theta.\)

#3. Express all the trigonometrical ratios in terms of \(cosecant.\)

Hint: You know where to start: \[
sin\theta = \frac{1}{cosec\theta}
\]

SL Loney – Examples 7

#2. If \(A = 45{°}\), verify that \(sin2A = 2sinAcosA\)

Solution:

\[
LHS = sin2A = sin90{°} = 1
\]

\[
RHS = 2sinAcosA = 2sin45{°}cos45{°}
\]\[
= 2 \times \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}} = 1
\]

#5. Verify that \(sin30{°}cos60{°} + cos30{°}sin60{°} = 1\)

Hint: Just put the values; you know them all.

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