Samacheer Kalvi Class 9 Maths Solutions Chapter 6 Trigonometry Exercise 6.3

Question 1. Find the value of the following:
(i) \( \left( \frac { \cos 47° }{ \sin 43° } \right)^2 + \left( \frac { \sin 72° }{ \cos 18° } \right)^2 - 2 \cos^2 45° \)
(ii) \( \frac { \cos 70° }{ \sin 20° } + \frac { \cos 59° }{ \sin 31° } + \frac { \cos \theta }{ \sin (90°- \theta) } - 8 \cos^2 60° \)
(iii) \( \tan 15° \tan 30° \tan 45° \tan 60° \tan 75° \)
(iv) \( \frac { \cot \theta }{ \tan (90°- \theta) } + \frac { \cos (90°- \theta) \tan \theta \sec (90°- \theta) }{ \sin (90°- \theta) \cot (90°- \theta) \operatorname{cosec} (90°- \theta) } \)
Answer:
(i) To find the value, we use trigonometric identities for complementary angles. We know that \( \cos(90° - \theta) = \sin \theta \) and \( \sin(90° - \theta) = \cos \theta \). So, \( \frac { \cos 47° }{ \sin 43° } = \frac { \cos (90° - 43°) }{ \sin 43° } = \frac { \sin 43° }{ \sin 43° } = 1 \). Similarly, \( \frac { \sin 72° }{ \cos 18° } = \frac { \sin (90° - 18°) }{ \cos 18° } = \frac { \cos 18° }{ \cos 18° } = 1 \). We also know that \( \cos 45° = \frac { 1 }{ \sqrt{2} } \). Now, substitute these values into the given expression: \( \left( \frac { \cos 47° }{ \sin 43° } \right)^2 + \left( \frac { \sin 72° }{ \cos 18° } \right)^2 - 2 \cos^2 45° \) \( = (1)^2 + (1)^2 - 2 \left( \frac { 1 }{ \sqrt{2} } \right)^2 \) \( = 1 + 1 - 2 \left( \frac { 1 }{ 2 } \right) \) \( = 2 - 1 \) \( = 1 \) The value of \( \cos 45° \) is a standard trigonometric value often used in these calculations.
(ii) For this part, we will use the same complementary angle identities as before. \( \frac { \cos 70° }{ \sin 20° } = \frac { \cos (90° - 20°) }{ \sin 20° } = \frac { \sin 20° }{ \sin 20° } = 1 \) \( \frac { \cos 59° }{ \sin 31° } = \frac { \cos (90° - 31°) }{ \sin 31° } = \frac { \sin 31° }{ \sin 31° } = 1 \) For the third term, \( \sin(90° - \theta) = \cos \theta \), so: \( \frac { \cos \theta }{ \sin (90°- \theta) } = \frac { \cos \theta }{ \cos \theta } = 1 \). We also know that \( \cos 60° = \frac { 1 }{ 2 } \). Substitute these values into the expression: \( \frac { \cos 70° }{ \sin 20° } + \frac { \cos 59° }{ \sin 31° } + \frac { \cos \theta }{ \sin (90°- \theta) } - 8 \cos^2 60° \) \( = 1 + 1 + 1 - 8 \left( \frac { 1 }{ 2 } \right)^2 \) \( = 3 - 8 \left( \frac { 1 }{ 4 } \right) \) \( = 3 - 2 \) \( = 1 \) It is important to remember the standard trigonometric values for common angles like \( 60° \).
(iii) First, write down the known standard values: \( \tan 30° = \frac { 1 }{ \sqrt{3} } \) \( \tan 45° = 1 \) \( \tan 60° = \sqrt{3} \) Substitute these values into the expression: \( \tan 15° \tan 30° \tan 45° \tan 60° \tan 75° \) \( = \tan 15° \times \frac { 1 }{ \sqrt{3} } \times 1 \times \sqrt{3} \times \tan 75° \) The terms \( \frac { 1 }{ \sqrt{3} } \) and \( \sqrt{3} \) cancel each other out: \( = \tan 15° \times \tan 75° \) Now, use the complementary angle identity \( \tan(90° - \theta) = \cot \theta \). So, \( \tan 75° = \tan(90° - 15°) = \cot 15° \). Also, remember that \( \cot \theta = \frac { 1 }{ \tan \theta } \). Substitute this back into the expression: \( = \tan 15° \times \cot 15° \) \( = \tan 15° \times \frac { 1 }{ \tan 15° } \) \( = 1 \) The product of tangent and cotangent of the same angle is always 1, which helps simplify the expression.
(iv) We will simplify each term using complementary angle identities and reciprocal identities. For the first term: We know \( \tan(90° - \theta) = \cot \theta \). So, \( \frac { \cot \theta }{ \tan (90°- \theta) } = \frac { \cot \theta }{ \cot \theta } = 1 \). For the second term, apply the complementary identities: \( \cos(90° - \theta) = \sin \theta \) \( \sec(90° - \theta) = \operatorname{cosec} \theta \) \( \sin(90° - \theta) = \cos \theta \) \( \cot(90° - \theta) = \tan \theta \) \( \operatorname{cosec}(90° - \theta) = \sec \theta \) Substitute these into the second term: \( \frac { \sin \theta \cdot \tan \theta \cdot \operatorname{cosec} \theta }{ \cos \theta \cdot \tan \theta \cdot \sec \theta } \) Now use the reciprocal identities \( \operatorname{cosec} \theta = \frac { 1 }{ \sin \theta } \) and \( \sec \theta = \frac { 1 }{ \cos \theta } \): \( \frac { \sin \theta \cdot \tan \theta \cdot \frac { 1 }{ \sin \theta } }{ \cos \theta \cdot \tan \theta \cdot \frac { 1 }{ \cos \theta } } \) Cancel out the \( \sin \theta \) terms in the numerator and \( \cos \theta \) terms in the denominator. Also, \( \tan \theta \) cancels from both numerator and denominator: \( = \frac { 1 }{ 1 } = 1 \) Finally, add the simplified first and second terms: \( = 1 + 1 \) \( = 2 \) Simplifying complex trigonometric expressions often involves converting all terms to sine and cosine, or using identities to cancel terms effectively.In simple words: For (i), (ii), and (iii), we used rules where angles adding up to 90 degrees simplify things, and also known values for certain angles. For (iv), we changed all the angles and trig functions using rules until everything cancelled out, giving us simple numbers.

🎯 Exam Tip: Always look for complementary angles (sum to 90°) to simplify expressions using identities like \( \sin(90° - \theta) = \cos \theta \) and \( \tan(90° - \theta) = \cot \theta \). Memorize standard trigonometric values for 30°, 45°, and 60°.