Let us determine the values of :sec245° – cot245° + sin230° – sin260°

⇒ sec45 = √2⇒ cot45 = 1⇒ ⇒ ⇒ sec245 – cot245 –sin230 –sin260= = = = 0

Q5 of 80 Page 302

Let us determine the values of :
sec²45° – cot²45° + sin²30° – sin²60°

⇒ sec45 = √2

⇒ cot45 = 1

⇒

⇒ sec²45 – cot²45 –sin²30 –sin²60

= 0

In a right angled triangle PQR, ∠Q=90°, ∠R=45°; if PR = 3√2, then let us find out the lengths of two sides PQ and QR.

Let us determine the values of :

sin²45° – cosec²60° + sec²30°

Let us determine the values of :

Questions · 80

23. Trigonometric Ratios and Trigonometric Identities

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