Question

Prove that: $(1 + \frac{1}{t a n ^{2} θ}) (1 + \frac{1}{c o t ^{2} θ}) = \frac{1}{s i n ^{2} θ - s i n ^{4} θ}$

Approach

Expand LHS using $cot^{2} θ = csc^{2} θ - 1$ and $tan^{2} θ = sec^{2} θ - 1$ . Simplify to get $\frac{1}{s i n ^{2} θ c o s ^{2} θ}$ . Use $cos^{2} θ = 1 - sin^{2} θ$ to match RHS.

Step-by-step working

1
LHS = $(1 + cot^{2} θ) (1 + tan^{2} θ) = csc^{2} θ \cdot sec^{2} θ$ .
1 mark
2
Rewrite: $\frac{1}{s i n ^{2} θ} \cdot \frac{1}{c o s ^{2} θ} = \frac{1}{s i n ^{2} θ c o s ^{2} θ}$ .
0.5 marks
3
Substitute $cos^{2} θ = 1 - sin^{2} θ$ : $\frac{1}{s i n ^{2} θ ( 1 - s i n ^{2} θ )}$ .
0.5 marks
4
Simplify: $\frac{1}{s i n ^{2} θ - s i n ^{4} θ}$ = RHS.
0.5 marks

Concepts used

trigonometric identityprovetancotsin

Approach

Step-by-step working

Concepts used

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