Q.1 Show that: 𝑐𝑜𝑠𝑒𝑐² θ − 𝑡𝑎𝑛² (90 − θ) = 𝑠𝑖𝑛² θ + 𝑠𝑖𝑛² (90   − θ)

 

(OR)

 

Q.2 Two tangents 𝑇𝑃 and 𝑇𝑄 are drawn to a circle with center 𝑂 from an external point 𝑇. Prove that ∠𝑃𝑇𝑄 = 2∠𝑂𝑃𝑄

We are given, 𝑐𝑜𝑠𝑒𝑐² θ − 𝑡𝑎𝑛² (90 − θ) = 𝑠𝑖𝑛² θ + 𝑠𝑖𝑛² (90   − θ)
We have to prove whether LHS is equal to RHS or not.
𝑐𝑜𝑡² θ = 𝑡𝑎𝑛(90 − θ)
𝑐𝑜𝑠² θ = 𝑠𝑖𝑛(90 − θ)
𝐿𝐻𝑆 = 𝑐𝑜𝑠𝑒𝑐² θ − 𝑡𝑎𝑛² (90 − θ)
⇒ 𝐿𝐻𝑆 = 𝑐𝑜𝑠𝑒𝑐² θ − (𝑡𝑎𝑛²(90 − θ))
⇒ 𝐿𝐻𝑆 = 𝑐𝑜𝑠𝑒𝑐² θ − (𝑐𝑜𝑡² θ)
⇒ 𝐿𝐻𝑆 = 𝑐𝑜𝑠𝑒𝑐² θ − 𝑐𝑜𝑡² θ
⇒ 𝐿𝐻𝑆 = 1
𝑅𝐻𝑆 = 𝑠𝑖𝑛² θ + 𝑠𝑖𝑛²(90 − θ)
⇒ 𝑅𝐻𝑆 = 𝑠𝑖𝑛² θ + (𝑠𝑖𝑛²(90 − θ))
⇒ 𝑅𝐻𝑆 = 𝑠𝑖𝑛² θ + (𝑐𝑜𝑠² θ)
⇒ 𝑅𝐻𝑆 = 𝑠𝑖𝑛² θ + 𝑐𝑜𝑠² θ
⇒ 𝑅𝐻𝑆 = 1
So, LHS = RHS.
Hence, proved.

(OR)

We are given two tangents, 𝑃𝑇 and 𝑇𝑄, and we have to prove that

∠𝑃𝑇𝑄 = 2∠𝑂𝑃𝑄

As we know that in a cyclic quadrilateral, that is, the quadrilateral inside a circle, the sum of opposite angles is 180$°$.
In the figure, we are given a cyclic quadrilateral 𝑂𝑄𝑃𝑅, in which the opposite angles are
∠𝑄𝑂𝑃 and ∠𝑃𝑇𝑄.
So, 𝑜 (as mentioned above)
⇒ ∠𝑄𝑂𝑃 = 180$°$ − ∠𝑃𝑇𝑄
A triangle 𝑂𝑃𝑄 will form if we join the points 𝑃 and 𝑄.
In triangle 𝑂𝑃𝑄, 𝑂𝑄 and 𝑂𝑃 are the radii of the circle, so, 𝑂𝑄 = 𝑂𝑃
As two sides of the ∆𝑂𝑃𝑄 are equal, this is an isosceles triangle.
In an isosceles triangle, the angles opposite to the equal sides are also equal, i.e.,
∠𝑂𝑃𝑄 = ∠𝑂𝑄𝑃
We know that the sum of the angles of a triangle is 180$°$
∆𝑂𝑃𝑄, ∠𝑃𝑂𝑄 + ∠𝑂𝑃𝑄 + ∠𝑂𝑄𝑃 = 180$°$
Putting the value of ∠𝑄𝑂𝑃 from eq(1), we get
180$°$ − ∠𝑃𝑇𝑄 + ∠𝑂𝑃𝑄 + ∠𝑂𝑃𝑄 = 180$°$ (as ∠𝑂𝑃𝑄 = ∠𝑂𝑄𝑃)
180$°$ will be cancelled from both sides,
⇒ 2∠𝑂𝑃𝑄 − ∠𝑃𝑇𝑄 = 180$°$
⇒ ∠𝑃𝑇𝑄 = 2∠𝑂𝑃𝑄
Hence, proved.