Question 1 iv Evaluate the following-iv sin 30-tan 45-cosec 60- 30-cos 60- 45-

(iv) (sin 30^∘ + tan 45^∘ cosec 60^∘)/(sec 30^∘ + cos 60^∘ + cot 45^∘) = (1/2+1 2/√(3))/(2/√(3)+1/2+1)= (3/2 2/√(3))/(3/2+2/√(3)) =3√(3) 4/2√(3)/3√(3) + 4/ ...

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Standard IX

Mathematics

Use of Trigonometric Table

Question 1 iv...

Question

Question 1 (iv)
Evaluate the following:
(iv) $\frac{(s i n 30^{\circ} + t a n 45^{\circ} - c o s e c 60^{\circ})}{(s e c 30^{\circ} + c o s 60^{\circ} + c o t 45^{\circ})}$

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Solution

(iv) $\frac{(s i n 30^{\circ} + t a n 45^{\circ} - c o s e c 60^{\circ})}{(s e c 30^{\circ} + c o s 60^{\circ} + c o t 45^{\circ})}$
$= \frac{(\frac{1}{2} + 1 - \frac{2}{\sqrt{3}})}{(\frac{2}{\sqrt{3}} + \frac{1}{2} + 1)}$ $= \frac{(\frac{3}{2} - \frac{2}{\sqrt{3}})}{(\frac{3}{2} + \frac{2}{\sqrt{3}})}$
$= \frac{\frac{3 \sqrt{3} - 4}{2 \sqrt{3}}}{\frac{3 \sqrt{3} + 4}{2 \sqrt{3}}}$ $= \frac{(3 \sqrt{3} - 4)}{(3 \sqrt{3} + 4)}$
$= \frac{(3 \sqrt{3} - 4) (3 \sqrt{3} - 4)}{(3 \sqrt{3} + 4) (3 \sqrt{3} - 4)}$ $= \frac{(3 \sqrt{3} - 4)^{2}}{(3 \sqrt{3})^{2} - (4)^{2}}$
$= \frac{(27 + 16 - 24 \sqrt{3})}{(27 - 16)}$ $= \frac{(43 - 24 \sqrt{3})}{11}$

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

Q. Question 1 (iv)
Evaluate the following:
(iv)

\frac{(s i n 30^{\circ} + t a n 45^{\circ} - c o s e c 60^{\circ})}{(s e c 30^{\circ} + c o s 60^{\circ} + c o t 45^{\circ})}

Evaluate the following

(i) sin60° cos30° + sin30° cos 60°

(ii) 2tan²45° + cos²30° − sin²60°

(iii)

(iv)

(v)

Solve : $\frac{sin 30^{\circ} + tan 45^{\circ} - c o s e c 60^{\circ}}{sec 30^{\circ} + cos 60^{\circ} + cot 45^{\circ}}$

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

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Use of Trigonometric Table

Standard IX Mathematics

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Question 1 (iv) Evaluate the following: (iv) (sin30∘+tan45∘−cosec60∘)(sec30∘+cos60∘+cot45∘)

(iv) (sin30∘+tan45∘−cosec60∘)(sec30∘+cos60∘+cot45∘) =(12+1−2√3)(2√3+12+1)=(32−2√3)(32+2√3) =3√3−42√33√3+42√3=(3√3−4)(3√3+4) =(3√3−4)(3√3−4)(3√3+4)(3√3−4)=(3√3−4)2(3√3)2−(4)2 =(27+16−24√3)(27−16)=(43−24√3)11

Question 1 (iv)
Evaluate the following:
(iv) $\frac{(s i n 30^{\circ} + t a n 45^{\circ} - c o s e c 60^{\circ})}{(s e c 30^{\circ} + c o s 60^{\circ} + c o t 45^{\circ})}$