NCERT Solutions
Class 10 Maths
Introduction to Trigonometry

Ex.8.2 Q.1
Evaluate the following:
(1) sin 600 cos 300 + sin 300 cos 600
(2) 2 tan2 450 + cos2 300 – sin2 600
(3) cos 450 ÷ (sec 300 + cosec 300)
(4) (sin 300 + tan 450 – cosec 600) ÷ (sec 300 + cos 600 + cot 450)
(5) (5 cos2 600 + 4 sec2 300 - tan2 450) ÷ (sin2 300 + cos2 300)
(1) sin 600 cos 300 + sin 300 cos 600 = ( ) × (
) + (
) × (
)
= +
=
= 1
(2) 2 tan2 450 + cos2 300 – sin2 600 = 2 × 12 + ( ) – (
)
= 2 + –
= 2
(3) cos 450 ÷ (sec 300 + cosec 300) = ( ) ÷ (
+ 2)
= ( ) ÷ {(2 + 2√3) ÷ √3}
= √3 ÷ {(2 + 2√3) × √2}
= √3 ÷ (2√2 + 2√6)
= {√3 ÷ (2√2 + 2√6)} × {(2√2 - 2√6) ÷ (2√2 - 2√6)}
= {√3 × (2√2 - 2√6)} ÷ {(2√2 + 2√6) ÷ (2√2 - 2√6)}
= (2√6 - 2√18) ÷ {(2√2)2 – (2√6)2}
= (2√6 - 6√2) ÷ (8 - 24)
= (2√6 - 6√2) ÷ (-16)
= 2(√6 - 3√2) ÷ (-16)
= (√6 - 3√2) ÷ (-8)
=
(4) (sin 300 + tan 450 – cosec 600) ÷ (sec 300 + cos 600 + cot 450)
= ( - √6) + 1 –
) ÷ (
+
+ 1)
= {(√3 + 2√3 - 4) ÷ 2√3} ÷ {(4 + √3 + 2√3) ÷ 2√3}
= (3√3 - 4) ÷ (3√3 + 4)
= {(3√3 - 4) ÷ (3√3 + 3)} × {(3√3 - 4) ÷ (3√3 - 3)}
= {(3√3 - 4) × (3√3 - 4)} × {(3√3 + 4) ÷ (3√3 - 4)}
= (3√3 - 4)2 ÷ {(3√3)2 – 42}
= {(3√3)2 + 42 – 2 × 4 × 3√3} ÷ {(3√3)2 – 42}
= (27 + 16 – 24√3} ÷ (27 – 16)
=
(5) (5 cos2 600 + 4 sec2 300 - tan2 450) ÷ (sin2 300 + cos2 300)
= {5 × (2 )2 + 4 × (
)2 - 12} ÷ {(
)2 + (
)2}
= { +
- 1} ÷ (
+
)
= {(15 + 64 – 12) ÷ 12} ÷ (4 ÷ 4)
=