Show that cos245- - -cos245-tan60- - -tan30-1

LHS=cos^2(45^∘ + θ)+cos^2(45^∘ θ)/tan(60^∘ + θ). tan(30^∘ θ) =cos^2(45^∘ + θ)+ [sin{90^∘ (45^∘ θ)} nbsp; ]^2/tan(60^∘+θ). cot{90^∘ (30^∘ θ)} [∵ sin ...

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Byju's Answer

Standard IX

Mathematics

Variation of Trigonometric Ratios from 0 to 90 Degrees

Show that cos...

Question

Show that $\frac{c o s^{2} (45^{\circ} + θ) + c o s^{2} (45^{\circ} - θ)}{t a n (60^{\circ} + θ) t a n (30^{\circ} - θ)} = 1$

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Solution

$L H S = \frac{c o s^{2} (45^{\circ} + θ) + c o s^{2} (45^{\circ} - θ)}{t a n (60^{\circ} + θ) . t a n (30^{\circ} - θ)}$

$= \frac{c o s^{2} (45^{\circ} + θ) + {[s i n {90^{\circ} - (45^{\circ} - θ)}]}^{2}}{t a n (60^{\circ} + θ) . c o t {90^{\circ} - (30^{\circ} - θ)}}$

$[∵ s i n (90^{\circ} - θ) = c o s θ a n d c o t (90^{\circ} - θ) = t a n θ]$

$= \frac{c o s^{2} (45^{\circ} + θ) + s i n^{2} (45^{\circ} + θ)}{t a n (60^{\circ} + θ) . c o t (60^{\circ} + θ)} [∵ s i n^{2} θ + c o s^{2} θ]$

$= \frac{1}{t a n (60^{\circ} + θ) \frac{1}{t a n (60^{\circ} + θ)}} = 1 = R H S$

Suggest Corrections

Show that cos2(45∘+θ)+cos2(45∘−θ)tan(60∘+θ)tan(30∘−θ)=1

Show that $\frac{c o s^{2} (45^{\circ} + θ) + c o s^{2} (45^{\circ} - θ)}{t a n (60^{\circ} + θ) t a n (30^{\circ} - θ)} = 1$