Show that the value of sin 45-sin 30-cos 45-cos 60- and 45-tan 45-cosec 45- 45- are equal

Sin 45^∘ sin 30^∘/cos 45^∘+cos space 60^∘ = 2 √(2)/2√(2)/2+√(2)/2√(2) = 2 √(2)/2+√(2) Multiplying numerator and denominator by sin 45^∘ tan 45^∘/cosec 4 ...

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Standard X

Mathematics

Range of Trigonometric Ratios from 0 to 90 Degrees

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Question

Show that the value of $\frac{s i n 45^{\circ} - s i n 30^{\circ}}{c o s 45^{\circ} + c o s 60^{\circ}}$ and $\frac{s e c 45^{\circ} - t a n 45^{\circ}}{c o s e c 45^{\circ} + c o t 45^{\circ}}$ are equal

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Solution

$\frac{s i n 45^{\circ} - s i n 30^{\circ}}{c o s 45^{\circ} + c o s s p a c e 60^{\circ}}$ = $\frac{\frac{2 - \sqrt{2}}{2 \sqrt{2}}}{\frac{2 + \sqrt{2}}{2 \sqrt{2}}}$ = $\frac{2 - \sqrt{2}}{2 + \sqrt{2}}$

Multiplying numerator and denominator by

$\frac{s i n 45^{\circ} - t a n 45^{\circ}}{c o s e c 45^{\circ} + c o t 45^{\circ}}$ = $\frac{2 - \sqrt{2}}{2 + \sqrt{2}}$

Hence the two are equal

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Similar questions

Q. Let

T_{1} = \frac{sin 45^{\circ} - sin 30^{\circ} + cot 90^{\circ}}{cos 45^{\circ} + cos 60^{\circ}}

and

T_{2} = \frac{sec 45^{\circ} - tan 45^{\circ}}{c o s e c 45^{\circ} + cos 0^{\circ} + cot 90^{\circ}}

, then

Q. The value of

(cos 0^{\circ} + sin 45^{\circ} + sin 30^{\circ}) (sin 90^{\circ} - cos 45^{\circ} + cos 60^{\circ})

Q. Find the value of

\frac{sin 30^{\circ} + tan 45^{\circ} - csc 60^{\circ}}{cot 45^{\circ} + cos 60^{\circ} - sec 30^{\circ}}

Q. The value of expression

\frac{sin 30^{\circ} + tan 45^{\circ} - sec 60^{\circ}}{cosec 30^{\circ} - cot 45^{\circ} - cos 60^{\circ}}

Q. Find the value of the following:

(cos 0^{\circ} + sin 45^{\circ} + sin 30^{\circ}) (sin 90^{\circ} - cos 45^{\circ} + cos 60^{\circ})

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Show that the value of sin 45∘−sin 30∘cos 45∘+cos 60∘ and sec 45∘−tan 45∘cosec 45∘+cot 45∘ are equal

sin 45∘−sin 30∘cos 45∘+cos space60∘ = 2−√22√22+√22√2 = 2−√22+√2 Multiplying numerator and denominator by sin 45∘−tan 45∘cosec 45∘+cot 45∘ = 2−√22+√2 Hence the two are equal

Show that the value of $\frac{s i n 45^{\circ} - s i n 30^{\circ}}{c o s 45^{\circ} + c o s 60^{\circ}}$ and $\frac{s e c 45^{\circ} - t a n 45^{\circ}}{c o s e c 45^{\circ} + c o t 45^{\circ}}$ are equal