Prove that sin θ sin( 90° - θ) - cos θ cos( 90° - θ) = 0 - Vidyadip

Question

Prove that sin θ sin( 90° - θ) - cos θ cos( 90° - θ) = 0

✓

Answer

LHS = sin θ sin( 90° - θ) - cos θ cos( 90° - θ)
= sin θ . cos θ - cos θ . sin θ
= 0
= RHS
Hence proved.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "[2 Mark Question Answer]" from Trigonometry→Trigonometry — all question types→Mathematics STD 10 question papers→ICSE Board STD 10 question papers→ICSE Board question paper generator→

Similar questions

The circumference of the base of a $12$ m high conical tent is $66$ m. Find the volume of the aircontained in it.

One card is drawn from a well shuffled deck of 52 cards. Find the probability of getting: the jack or the queen of the hearts

Find the fourth proportional to $\frac{1}{3}, \frac{1}{4}, \frac{1}{5}$

Without actually determining the roots comment upon the nature of the roots of each of the following equations:
$3x^2 + 2x - 1 = 0$

Describe the locus for questions 1 to 13 given below:

The locus of the centre of a wheel of a bicycle going straight along a level road.

A card is drawn from a pack of well - shuffled 52 playing cards . Find the probability that the card drawn is
(iv) an ace or a queen

Two dice (each bearing numbers 1 to 6) are rolled together. Find the probability that the sum of the numbers on the upper-most faces of two dice is: between 5 and 8

Solve the following equation:
$(2x+3) (3x-7) = 0$

What number must be added to $2x^3 – 7x^2 + 2x$ so that the resulting polynomial leaves the remainder – 2 when divided by $2x – 3?$

When divided by $x-3$ the polynomials $x^3-p x^2+x+6$ and $2 x^3-x^2-(p+3) x-6$ leave the same remainder. Find the value of 'p'.