Applications and Mixed Problems

Pulling the Tools Together

This section blends everything: the six ratios, the standard-angle table, the three identities, and complementary angles. Most exam questions combine two or three of these ideas, so the skill is recognising which tool fits.

A quick decision guide:

• A numerical expression in standard angles ⇒ use the table.
• 'Given one ratio, find another' ⇒ build the triangle or use an identity.
• Angles that sum to 90° ⇒ use complementary relations.
• 'Prove that …' ⇒ convert to sin/cos and use $\sin^2+\cos^2=1$.

Key Point: Read the question, spot the structure, then pick the matching tool.

[Board Important] Showing the method clearly earns marks even when the final value is a messy surd.

Finding Ratios from One Given Ratio

When one ratio is given, two routes work:

1. Triangle route: assign sides matching the ratio, use Pythagoras for the third side, read off the rest.
2. Identity route: use $\sin^2+\cos^2=1$, $\sec^2-\tan^2=1$, or $\csc^2-\cot^2=1$.

For acute angles, all ratios are positive, so you take positive square roots.

Key Point: The triangle route is more intuitive; the identity route is faster once you're fluent.

[JEE/NEET Tip] For acute angles you never worry about signs — every ratio is positive. (Signs matter only in Class 11 when angles exceed 90°.)

Evaluating Composite Expressions

For a long expression, evaluate term by term:

• Replace each standard-angle ratio with its value.
• Simplify squares and products before adding.
• Use complementary relations to pair and cancel terms.

Keep surds like $\sqrt2, \sqrt3$ exact; rationalise only at the end if needed.

Key Point: Term-by-term substitution, then simplify — don't try to do it all in one line.

[Board Important] A common slip is mis-squaring: $(\tfrac{\sqrt3}{2})^2 = \tfrac34$, not $\tfrac{\sqrt3}{4}$.

A Glimpse of Heights and Distances

The next chapter (Applications of Trigonometry) uses these ratios in real-world problems with an angle of elevation (looking up) or angle of depression (looking down). The core idea: model the situation as a right triangle, then apply $\sin$, $\cos$, or $\tan$.

For example, if a tower of height $h$ casts a shadow of length $\ell$ and the sun's elevation is $\theta$, then $\tan\theta = \dfrac{h}{\ell}$.

Key Point: Real-world height/distance problems reduce to choosing the ratio that links the known and unknown sides.

[Board Important] Mastering the ratios here makes the next chapter almost mechanical.

Solved Examples

Example 1: Given tan, evaluate

If $\tan\theta = \tfrac34$, evaluate $\dfrac{1 - \sin\theta}{1 + \sin\theta}$.

Solution:

1. With Opp 3, Adj 4, Hyp 5: $\sin\theta = \tfrac35$.
2. $\dfrac{1 - 3/5}{1 + 3/5} = \dfrac{2/5}{8/5} = \dfrac14$.

Final Answer: $\tfrac14$.

Takeaway: Build the triangle, then substitute.

Example 2: Standard-angle expression

Evaluate $\dfrac{5\cos^2 60° + 4\sec^2 30° - \tan^2 45°}{\sin^2 30° + \cos^2 30°}$.

Solution:

1. Numerator: $5(\tfrac12)^2 + 4(\tfrac{2}{\sqrt3})^2 - 1^2 = 5\cdot\tfrac14 + 4\cdot\tfrac43 - 1 = \tfrac54 + \tfrac{16}{3} - 1$.
2. $= \tfrac{15}{12} + \tfrac{64}{12} - \tfrac{12}{12} = \tfrac{67}{12}$.
3. Denominator $= 1$.

Final Answer: $\tfrac{67}{12}$.

Takeaway: Evaluate numerator term by term; the denominator is the identity = 1.

Example 3: Identity + value

If $\cos\theta = \tfrac{7}{25}$, find $\dfrac{\sin\theta - \cos\theta}{\sin\theta + \cos\theta}$.

Solution:

1. $\sin\theta = \sqrt{1 - (7/25)^2} = \tfrac{24}{25}$.
2. $\dfrac{24/25 - 7/25}{24/25 + 7/25} = \dfrac{17}{31}$.

Final Answer: $\tfrac{17}{31}$.

Takeaway: Get sin from cos, then substitute.

Example 4: Mixed with complementary

Evaluate $\dfrac{\tan 65°}{\cot 25°}$.

Solution:

1. $\cot 25° = \cot(90° - 65°) = \tan 65°$.
2. Ratio $= 1$.

Final Answer: 1.

Takeaway: $65° + 25° = 90°$, and $\tan$ pairs with $\cot$.

Example 5: Prove and evaluate

If $3\tan\theta = 4$, find $\dfrac{3\sin\theta + 2\cos\theta}{3\sin\theta - 2\cos\theta}$.

Solution:

1. $\tan\theta = \tfrac43$. Divide top and bottom by $\cos\theta$: $\dfrac{3\tan\theta + 2}{3\tan\theta - 2}$.
2. $= \dfrac{3\cdot\tfrac43 + 2}{3\cdot\tfrac43 - 2} = \dfrac{4 + 2}{4 - 2} = \dfrac{6}{2} = 3$.

Final Answer: 3.

Takeaway: Dividing by $\cos\theta$ turns the expression into one in $\tan\theta$.

Example 6: Height application

A tower casts a shadow 15 m long when the sun's angle of elevation is 60°. Find the tower's height.

Solution:

1. $\tan 60° = \dfrac{h}{15} \Rightarrow h = 15\tan 60° = 15\sqrt3$.

Final Answer: $15\sqrt3 \approx 25.98$ m.

Takeaway: $\tan(\text{elevation}) = \dfrac{\text{height}}{\text{shadow}}$.

Example 7: Combine identities

If $\sin\theta + \cos\theta = p$ and $\sec\theta + \csc\theta = q$, show $q(p^2 - 1) = 2p$.

Solution:

1. $p^2 - 1 = (\sin\theta+\cos\theta)^2 - 1 = 2\sin\theta\cos\theta$.
2. $q = \dfrac{1}{\cos\theta} + \dfrac{1}{\sin\theta} = \dfrac{\sin\theta + \cos\theta}{\sin\theta\cos\theta} = \dfrac{p}{\sin\theta\cos\theta}$.
3. $q(p^2 - 1) = \dfrac{p}{\sin\theta\cos\theta}\cdot 2\sin\theta\cos\theta = 2p$.

Final Answer: Proved.

Takeaway: Express both $p^2-1$ and $q$ in terms of $\sin\theta\cos\theta$.

Example 8: Eliminate theta

If $x = r\sin\theta\cos\phi$, $y = r\sin\theta\sin\phi$, $z = r\cos\theta$, show $x^2 + y^2 + z^2 = r^2$.

Solution:

1. $x^2 + y^2 = r^2\sin^2\theta(\cos^2\phi + \sin^2\phi) = r^2\sin^2\theta$.
2. Add $z^2 = r^2\cos^2\theta$: $r^2\sin^2\theta + r^2\cos^2\theta = r^2$.

Final Answer: Proved.

Takeaway: Apply $\sin^2+\cos^2=1$ twice (once in $\phi$, once in $\theta$).

Example 9: Value with given sec

If $\sec\theta + \tan\theta = 2$, find $\sec\theta - \tan\theta$ and hence $\sec\theta$.

Solution:

1. $(\sec\theta + \tan\theta)(\sec\theta - \tan\theta) = \sec^2\theta - \tan^2\theta = 1$.
2. So $\sec\theta - \tan\theta = \tfrac12$.
3. Adding: $2\sec\theta = 2 + \tfrac12 = \tfrac52 \Rightarrow \sec\theta = \tfrac54$.

Final Answer: $\sec\theta - \tan\theta = \tfrac12$, $\sec\theta = \tfrac54$.

Takeaway: Conjugate product equals 1; add the two equations to isolate sec.

Example 10: Full mixed evaluation

Evaluate $\dfrac{\sin 18°}{\cos 72°} + \sqrt3\,\tan 10° \tan 30° \tan 80°$.

Solution:

1. $\cos 72° = \sin 18°$, so first term $= 1$.
2. $\tan 80° = \cot 10°$, so $\tan 10°\tan 80° = 1$.
3. $\sqrt3\,\tan 30° = \sqrt3\cdot\tfrac{1}{\sqrt3} = 1$.
4. Second term $= 1\cdot 1 = 1$.
5. Total $= 1 + 1 = 2$.

Final Answer: 2.

Takeaway: Spot complementary pairs ($18°,72°$ and $10°,80°$) to collapse the expression.